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Tag 00PK

Chapter 10: Commutative Algebra > Section 10.121: Quasi-finite maps

Equivalent conditions for isolated points in fibres

Lemma 10.121.2. Let $R \to S$ be a ring map of finite type. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Let $F = \mathop{\rm Spec}(S \otimes_R \kappa(\mathfrak p))$ be the fibre of $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$, see Remark 10.16.8. Denote $\overline{\mathfrak q} \in F$ the point corresponding to $\mathfrak q$. The following are equivalent

  1. $\overline{\mathfrak q}$ is an isolated point of $F$,
  2. $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is finite over $\kappa(\mathfrak p)$,
  3. there exists a $g \in S$, $g \not \in \mathfrak q$ such that the only prime of $D(g)$ mapping to $\mathfrak p$ is $\mathfrak q$,
  4. $\dim_{\overline{\mathfrak q}}(F) = 0$,
  5. $\overline{\mathfrak q}$ is a closed point of $F$ and $\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$, and
  6. the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ is finite and $\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$.

Proof. Note that $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} = (S \otimes_R \kappa(\mathfrak p))_{\overline{\mathfrak q}}$. Moreover $S \otimes_R \kappa(\mathfrak p)$ is of finite type over $\kappa(\mathfrak p)$. The conditions correspond exactly to the conditions of Lemma 10.121.1 for the $\kappa(\mathfrak p)$-algebra $S \otimes_R \kappa(\mathfrak p)$ and the prime $\overline{\mathfrak q}$, hence they are equivalent. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 28877–28901 (see updates for more information).

    \begin{lemma}
    \label{lemma-isolated-point-fibre}
    \begin{slogan}
    Equivalent conditions for isolated points in fibres
    \end{slogan}
    Let $R \to S$ be a ring map of finite type.
    Let $\mathfrak q \subset S$ be a prime lying over
    $\mathfrak p \subset R$. Let $F = \Spec(S \otimes_R \kappa(\mathfrak p))$
    be the fibre of $\Spec(S) \to \Spec(R)$, see
    Remark \ref{remark-fundamental-diagram}.
    Denote $\overline{\mathfrak q} \in F$ the point corresponding to
    $\mathfrak q$. The following are equivalent
    \begin{enumerate}
    \item $\overline{\mathfrak q}$ is an isolated point of $F$,
    \item $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is finite over
    $\kappa(\mathfrak p)$,
    \item there exists a $g \in S$, $g \not \in \mathfrak q$ such that
    the only prime of $D(g)$ mapping to $\mathfrak p$ is $\mathfrak q$,
    \item $\dim_{\overline{\mathfrak q}}(F) = 0$,
    \item $\overline{\mathfrak q}$ is a closed point of $F$ and
    $\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$, and
    \item the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$
    is finite and $\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Note that $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} =
    (S \otimes_R \kappa(\mathfrak p))_{\overline{\mathfrak q}}$.
    Moreover $S \otimes_R \kappa(\mathfrak p)$ is of finite type over
    $\kappa(\mathfrak p)$.
    The conditions correspond exactly to the conditions of
    Lemma \ref{lemma-isolated-point}
    for the $\kappa(\mathfrak p)$-algebra $S \otimes_R \kappa(\mathfrak p)$
    and the prime $\overline{\mathfrak q}$, hence they are equivalent.
    \end{proof}

    Comments (1)

    Comment #1654 by Matthieu Romagny on September 24, 2015 a 9:11 pm UTC

    Suggested slogan: Equivalent conditions for isolated points in fibres

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