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Tag 00U6

Chapter 10: Commutative Algebra > Section 10.141: Étale ring maps

Lemma 10.141.7. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over a prime $\mathfrak p$ of $R$. If

  1. $R \to S$ is of finite presentation,
  2. $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat
  3. $\mathfrak p S_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and
  4. the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ is finite separable,

then $R \to S$ is étale at $\mathfrak q$.

Proof. Apply Lemma 10.121.2 to find a $g \in S$, $g \not \in \mathfrak q$ such that $\mathfrak q$ is the only prime of $S_g$ lying over $\mathfrak p$. We may and do replace $S$ by $S_g$. Then $S \otimes_R \kappa(\mathfrak p)$ has a unique prime, hence is a local ring, hence is equal to $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} \cong \kappa(\mathfrak q)$. By Lemma 10.135.16 there exists a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is smooth. Replace $S$ by $S_g$ again we may assume that $R \to S$ is smooth. By Lemma 10.135.10 we may even assume that $R \to S$ is standard smooth, say $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$. Since $S \otimes_R \kappa(\mathfrak p) = \kappa(\mathfrak q)$ has dimension $0$ we conclude that $n = c$, i.e., if $R \to S$ is étale. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 37543–37556 (see updates for more information).

    \begin{lemma}
    \label{lemma-characterize-etale}
    Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$
    lying over a prime $\mathfrak p$ of $R$. If
    \begin{enumerate}
    \item $R \to S$ is of finite presentation,
    \item $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat
    \item $\mathfrak p S_{\mathfrak q}$ is the maximal ideal
    of the local ring $S_{\mathfrak q}$, and
    \item the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$
    is finite separable,
    \end{enumerate}
    then $R \to S$ is \'etale at $\mathfrak q$.
    \end{lemma}
    
    \begin{proof}
    Apply
    Lemma \ref{lemma-isolated-point-fibre}
    to find a $g \in S$, $g \not \in \mathfrak q$ such that
    $\mathfrak q$ is the only prime of $S_g$ lying over $\mathfrak p$.
    We may and do replace $S$ by $S_g$. Then
    $S \otimes_R \kappa(\mathfrak p)$ has a unique prime, hence is a
    local ring, hence is equal to
    $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}
    \cong \kappa(\mathfrak q)$.
    By Lemma \ref{lemma-flat-fibre-smooth}
    there exists a $g \in S$, $g \not \in \mathfrak q$
    such that $R \to S_g$ is smooth. Replace $S$ by $S_g$ again we may
    assume that $R \to S$ is smooth. By
    Lemma \ref{lemma-smooth-syntomic} we may even assume that
    $R \to S$ is standard smooth, say $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$.
    Since $S \otimes_R \kappa(\mathfrak p) = \kappa(\mathfrak q)$
    has dimension $0$ we conclude that $n = c$, i.e.,
    if $R \to S$ is \'etale.
    \end{proof}

    Comments (2)

    Comment #3272 by Dario Weißmann on April 11, 2018 a 5:10 pm UTC

    Last sentence: 'i.e., if $R\to S$ is étale.'

    Comment #3278 by Luo Wenbin on April 16, 2018 a 1:10 am UTC

    Suggested slogan: It’s about a criterion to determine that a ring map is etale

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