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Tag 01TE

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

Lemma 28.19.2. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Set $s = f(x)$. If $\kappa(x)/\kappa(s)$ is an algebraic field extension, then

  1. $x$ is a closed point of its fibre, and
  2. if in addition $s$ is a closed point of $S$, then $x$ is a closed point of $X$.

Proof. The second statement follows from the first by elementary topology. According to Schemes, Lemma 25.18.5 to prove the first statement we may replace $X$ by $X_s$ and $S$ by $\mathop{\rm Spec}(\kappa(s))$. Thus we may assume that $S = \mathop{\rm Spec}(k)$ is the spectrum of a field. In this case, let $\mathop{\rm Spec}(A) = U \subset X$ be any affine open containing $x$. The point $x$ corresponds to a prime ideal $\mathfrak q \subset A$ such that $k \subset \kappa(\mathfrak q)$ is an algebraic field extension. By Algebra, Lemma 10.34.9 we see that $\mathfrak q$ is a maximal ideal, i.e., $x \in U$ is a closed point. Since the affine opens form a basis of the topology of $X$ we conclude that $\{x\}$ is closed. $\square$

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 3153–3164 (see updates for more information).

    \begin{lemma}
    \label{lemma-algebraic-residue-field-extension-closed-point-fibre}
    Let $f : X \to S$ be a morphism of schemes.
    Let $x \in X$ be a point. Set $s = f(x)$.
    If $\kappa(x)/\kappa(s)$
    is an algebraic field extension, then
    \begin{enumerate}
    \item $x$ is a closed point of its fibre, and
    \item if in addition $s$ is a closed point of $S$, then
    $x$ is a closed point of $X$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    The second statement follows from the first by elementary topology.
    According to Schemes, Lemma \ref{schemes-lemma-fibre-topological}
    to prove the first statement
    we may replace $X$ by $X_s$ and $S$ by $\Spec(\kappa(s))$.
    Thus we may assume that $S = \Spec(k)$ is the spectrum of a field.
    In this case, let $\Spec(A) = U \subset X$ be any affine open
    containing $x$. The point $x$ corresponds to a prime ideal
    $\mathfrak q \subset A$ such that $k \subset \kappa(\mathfrak q)$
    is an algebraic field extension. By
    Algebra, Lemma \ref{algebra-lemma-finite-residue-extension-closed}
    we see that $\mathfrak q$ is a maximal ideal, i.e., $x \in U$ is a
    closed point. Since the affine opens form
    a basis of the topology of $X$ we conclude that $\{x\}$ is closed.
    \end{proof}

    Comments (2)

    Comment #2265 by yangan on November 1, 2016 a 3:26 am UTC

    In the second line, the inclusion is reversed, should be k(s)\subset k(x).

    Comment #2294 by Johan (site) on November 3, 2016 a 11:34 pm UTC

    Actually, I have been trying to use consistently the notation $K/L$ for field extensions. So I fixed it that way. Thanks!

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