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 10.44.2.

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. $\square$

Comment #8549 by on

In the second paragraph of the proof, instead of Lemma 00TG, it seems more relevant to cite Lemma 00TF.

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