The Stacks project

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). $\square$


Comments (2)

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 ?


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02FM. Beware of the difference between the letter 'O' and the digit '0'.