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., $R \to S$ is étale.
$\square$

## Comments (3)

Comment #3256 by Dario Weißmann on

Comment #3262 by Luo Wenbin on

Comment #3352 by Johan on