Lemma 10.143.7. 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 $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is finite separable,
then $R \to S$ is étale at $\mathfrak q$.
Proof. Apply Lemma 10.122.2 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 10.137.17 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 10.137.10 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., $R \to S$ is étale. $\square$
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