## 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

- $R \to S$ is of finite presentation,
- $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat
- $\mathfrak p S_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and
- 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}
```

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