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

1. $R \to S$ is of finite type,

2. $\mathfrak p S_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and

3. the field extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ 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). $\square$

Comment #4876 by HAO on

Sorry but I can't see why Lemma 10.121.1 is needed in the proof of Lemma 10.148.7. Where did you use $S=S'\times k(q)$?

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).