Lemma 10.140.5 (00TV)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.140: Smooth algebras over fields
Lemma 10.140.5 (cite)

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Lemma 10.140.5. Let $k$ be a field. Let $S$ be a finite type $k$ -algebra. Let $\mathfrak q \subset S$ be a prime. Assume $\kappa (\mathfrak q)$ is separable over $k$ . The following are equivalent:

The algebra $S$ is smooth at $\mathfrak q$ over $k$ .
The ring $S_{\mathfrak q}$ is regular.

Proof. Denote $R = S_{\mathfrak q}$ and denote its maximal by $\mathfrak m$ and its residue field $\kappa$ . By Lemma 10.140.4 and 10.131.9 we see that there is a short exact sequence

$0 \to \mathfrak m/\mathfrak m^2 \to \Omega _{R/k} \otimes _ R \kappa \to \Omega _{\kappa /k} \to 0$

Note that $\Omega _{R/k} = \Omega _{S/k, \mathfrak q}$ , see Lemma 10.131.8. Moreover, since $\kappa$ is separable over $k$ we have $\dim _{\kappa } \Omega _{\kappa /k} = \text{trdeg}_ k(\kappa )$ . Hence we get

$\dim _{\kappa } \Omega _{R/k} \otimes _ R \kappa = \dim _\kappa \mathfrak m/\mathfrak m^2 + \text{trdeg}_ k (\kappa ) \geq \dim R + \text{trdeg}_ k (\kappa ) = \dim _{\mathfrak q} S$

(see Lemma 10.116.3 for the last equality) with equality if and only if $R$ is regular. Thus we win by applying Lemma 10.140.3. $\square$

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