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Tag 0BAJ

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

Lemma 28.48.6. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be a birational morphism between schemes which have finitely many irreducible components. Assume

  1. either $f$ is quasi-compact or $f$ is separated, and
  2. either $f$ is locally of finite type and $Y$ is reduced or $f$ is locally of finite presentation.

Then there exists a dense open $V \subset Y$ such that $f^{-1}(V) \to V$ is an isomorphism.

Proof. By Lemma 28.48.5 we may assume that $f$ is finite. Since $Y$ has finitely many irreducible components, we can find a dense open which is a disjoint union of its irreducible components. Thus we may assume $Y$ is irreducible. By Lemma 28.47.5 we find a nonempty open $U \subset X$ such that $f|_U : U \to Y$ is an open immersion. After removing the closed (as $f$ finite) subset $f(X \setminus U)$ from $Y$ we see that $f$ is an isomorphism. $\square$

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 12031–12043 (see updates for more information).

    \begin{lemma}
    \label{lemma-birational-isomorphism-over-dense-open}
    Let $X$, $Y$ be schemes. Let $f : X \to Y$ be a birational morphism
    between schemes which have finitely many irreducible components.
    Assume
    \begin{enumerate}
    \item either $f$ is quasi-compact or $f$ is separated, and
    \item either $f$ is locally of finite type and $Y$ is reduced or
    $f$ is locally of finite presentation.
    \end{enumerate}
    Then there exists a dense open $V \subset Y$
    such that $f^{-1}(V) \to V$ is an isomorphism.
    \end{lemma}
    
    \begin{proof}
    By Lemma \ref{lemma-finite-over-dense-open} we may assume that $f$ is finite.
    Since $Y$ has finitely many irreducible components, we can find a dense
    open which is a disjoint union of its irreducible components. Thus we may
    assume $Y$ is irreducible. By Lemma \ref{lemma-birational-birational} we find
    a nonempty open $U \subset X$ such that $f|_U : U \to Y$ is an open immersion.
    After removing the closed (as $f$ finite)
    subset $f(X \setminus U)$ from $Y$ we see that $f$ is an isomorphism.
    \end{proof}

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