Lemma 10.151.7. 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 type,
$\mathfrak p S_{\mathfrak q}$ 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 unramified at $\mathfrak q$.
Proof. By Lemma 10.151.3 (8) it suffices to show that $\Omega _{S \otimes _ R \kappa (\mathfrak p) / \kappa (\mathfrak p)}$ is zero when localized at $\mathfrak q$. Hence we may replace $S$ by $S \otimes _ R \kappa (\mathfrak p)$ and $R$ by $\kappa (\mathfrak p)$. In other words, we may assume that $R = k$ is a field and $S$ is a finite type $k$-algebra. In this case the hypotheses imply that $S_{\mathfrak q} \cong \kappa (\mathfrak q)$. Thus $(\Omega _{S/k})_{\mathfrak q} = \Omega _{S_\mathfrak q/k} = \Omega _{\kappa (\mathfrak q)/k}$ is zero as desired (the first equality is Lemma 10.131.8). $\square$
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