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Tag 02NW

Chapter 28: Morphisms of Schemes > Section 28.48: Generically finite morphisms

Lemma 28.48.1. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be locally of finite type. Let $\eta \in Y$ be a generic point of an irreducible component of $Y$. The following are equivalent:

  1. the set $f^{-1}(\{\eta\})$ is finite,
  2. there exist affine opens $U_i \subset X$, $i = 1, \ldots, n$ and $V \subset Y$ with $f(U_i) \subset V$, $\eta \in V$ and $f^{-1}(\{\eta\}) \subset \bigcup U_i$ such that each $f|_{U_i} : U_i \to V$ is finite.

If $f$ is quasi-separated, then these are also equivalent to

  1. (3)    there exist affine opens $V \subset Y$, and $U \subset X$ with $f(U) \subset V$, $\eta \in V$ and $f^{-1}(\{\eta\}) \subset U$ such that $f|_U : U \to V$ is finite.

If $f$ is quasi-compact and quasi-separated, then these are also equivalent to

  1. (4)    there exists an affine open $V \subset Y$, $\eta \in V$ such that $f^{-1}(V) \to V$ is finite.

Proof. The question is local on the base. Hence we may replace $Y$ by an affine neighbourhood of $\eta$, and we may and do assume throughout the proof below that $Y$ is affine, say $Y = \mathop{\rm Spec}(R)$.

It is clear that (2) implies (1). Assume that $f^{-1}(\{\eta\}) = \{\xi_1, \ldots, \xi_n\}$ is finite. Choose affine opens $U_i \subset X$ with $\xi_i \in U_i$. By Algebra, Lemma 10.121.10 we see that after replacing $Y$ by a standard open in $Y$ each of the morphisms $U_i \to Y$ is finite. In other words (2) holds.

It is clear that (3) implies (1). Assume $f^{-1}(\{\eta\}) = \{\xi_1, \ldots, \xi_n\}$ and assume that $f$ is quasi-separated. Since $Y$ is affine this implies that $X$ is quasi-separated. Since each $\xi_i$ maps to a generic point of an irreducible component of $Y$, we see that each $\xi_i$ is a generic point of an irreducible component of $X$. By Properties, Lemma 27.29.1 we can find an affine open $U \subset X$ containing each $\xi_i$. By Algebra, Lemma 10.121.10 we see that after replacing $Y$ by a standard open in $Y$ the morphisms $U \to Y$ is finite. In other words (3) holds.

It is clear that (4) implies all of (1) – (3) with no further assumptions on $f$. Suppose that $f$ is quasi-compact and quasi-separated. We have to show that the equivalent conditions (1) – (3) imply (4). Let $U$, $V$ be as in (3). Replace $Y$ by $V$. Since $f$ is quasi-compact and $Y$ is quasi-compact (being affine) we see that $X$ is quasi-compact. Hence $Z = X \setminus U$ is quasi-compact, hence the morphism $f|_Z : Z \to Y$ is quasi-compact. By construction of $Z$ we see that $\eta \not \in f(Z)$. Hence by Lemma 28.8.4 we see that there exists an affine open neighbourhood $V'$ of $\eta$ in $Y$ such that $f^{-1}(V') \cap Z = \emptyset$. Then we have $f^{-1}(V') \subset U$ and this means that $f^{-1}(V') \to V'$ is finite. $\square$

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 11801–11827 (see updates for more information).

    \begin{lemma}
    \label{lemma-generically-finite}
    Let $X$, $Y$ be schemes.
    Let $f : X \to Y$ be locally of finite type.
    Let $\eta \in Y$ be a generic point of an irreducible component
    of $Y$. The following are equivalent:
    \begin{enumerate}
    \item the set $f^{-1}(\{\eta\})$ is finite,
    \item there exist affine opens $U_i \subset X$, $i = 1, \ldots, n$
    and $V \subset Y$ with $f(U_i) \subset V$,
    $\eta \in V$ and $f^{-1}(\{\eta\}) \subset \bigcup U_i$
    such that each $f|_{U_i} : U_i \to V$ is finite.
    \end{enumerate}
    If $f$ is quasi-separated, then these are also equivalent to
    \begin{enumerate}
    \item[(3)] there exist affine opens $V \subset Y$,
    and $U \subset X$ with $f(U) \subset V$,
    $\eta \in V$ and $f^{-1}(\{\eta\}) \subset U$
    such that $f|_U : U \to V$ is finite.
    \end{enumerate}
    If $f$ is quasi-compact and quasi-separated,
    then these are also equivalent to
    \begin{enumerate}
    \item[(4)] there exists an affine open $V \subset Y$, $\eta \in V$
    such that $f^{-1}(V) \to V$ is finite.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    The question is local on the base. Hence we may replace $Y$ by an
    affine neighbourhood of $\eta$, and we may and do assume throughout
    the proof below that $Y$ is affine, say $Y = \Spec(R)$.
    
    \medskip\noindent
    It is clear that (2) implies (1).
    Assume that $f^{-1}(\{\eta\}) = \{\xi_1, \ldots, \xi_n\}$ is finite.
    Choose affine opens $U_i \subset X$ with $\xi_i \in U_i$.
    By Algebra, Lemma \ref{algebra-lemma-generically-finite} we see
    that after replacing $Y$ by a standard open in
    $Y$ each of the morphisms $U_i \to Y$ is finite.
    In other words (2) holds.
    
    \medskip\noindent
    It is clear that (3) implies (1).
    Assume $f^{-1}(\{\eta\}) = \{\xi_1, \ldots, \xi_n\}$ and assume
    that $f$ is quasi-separated.
    Since $Y$ is affine this implies that $X$ is quasi-separated.
    Since each $\xi_i$ maps to a generic point of an irreducible component
    of $Y$, we see that each $\xi_i$ is a generic point of an irreducible
    component of $X$.
    By Properties, Lemma \ref{properties-lemma-maximal-points-affine}
    we can find an affine open $U \subset X$ containing each $\xi_i$.
    By Algebra, Lemma \ref{algebra-lemma-generically-finite} we see
    that after replacing $Y$ by a standard open in
    $Y$ the morphisms $U \to Y$ is finite.
    In other words (3) holds.
    
    \medskip\noindent
    It is clear that (4) implies all of (1) -- (3) with no further assumptions
    on $f$. Suppose that $f$ is quasi-compact and quasi-separated. We have to
    show that the equivalent conditions (1) -- (3) imply (4).
    Let $U$, $V$ be as in (3). Replace $Y$ by $V$. Since $f$ is quasi-compact
    and $Y$ is quasi-compact (being affine) we see that $X$ is quasi-compact.
    Hence $Z = X \setminus U$ is quasi-compact, hence the morphism
    $f|_Z : Z \to Y$ is quasi-compact. By construction of $Z$ we see that
    $\eta \not \in f(Z)$. Hence by
    Lemma \ref{lemma-quasi-compact-generic-point-not-in-image}
    we see that there exists an affine open
    neighbourhood $V'$ of $\eta$ in $Y$ such that $f^{-1}(V') \cap Z = \emptyset$.
    Then we have $f^{-1}(V') \subset U$ and this means
    that $f^{-1}(V') \to V'$ is finite.
    \end{proof}

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