Lemma 10.140.9. Let $R \to S$ be an injective finite type ring map with $R$ and $S$ domains. Then $R \to S$ is smooth at $\mathfrak q = (0)$ if and only if the induced extension $L/K$ of fraction fields is separable.
Proof. Assume $R \to S$ is smooth at $(0)$. We may replace $S$ by $S_ g$ for some nonzero $g \in S$ and assume that $R \to S$ is smooth. Then $K \to S \otimes _ R K$ is smooth (Lemma 10.137.4). Moreover, for any field extension $K'/K$ the ring map $K' \to S \otimes _ R K'$ is smooth as well. Hence $S \otimes _ R K'$ is a regular ring by Lemma 10.140.3, in particular reduced. It follows that $S \otimes _ R K$ is a geometrically reduced over $K$. Hence $L$ is geometrically reduced over $K$, see Lemma 10.43.3. Hence $L/K$ is separable by Lemma 10.44.1.
Conversely, assume that $L/K$ is separable. We may assume $R \to S$ is of finite presentation, see Lemma 10.30.1. It suffices to prove that $K \to S \otimes _ R K$ is smooth at $(0)$, see Lemma 10.137.18. This follows from Lemma 10.140.5, the fact that a field is a regular ring, and the assumption that $L/K$ is separable. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (1)
Comment #8549 by Manolis Tsakiris on