Lemma 10.140.1 (00TR)—The Stacks project

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Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.140: Smooth algebras over fields
Lemma 10.140.1 (cite)

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Lemma 10.140.1. Let $k$ be an algebraically closed field. Let $S$ be a finite type $k$ -algebra. Let $\mathfrak m \subset S$ be a maximal ideal. Then

$\dim _{\kappa (\mathfrak m)} \Omega _{S/k} \otimes _ S \kappa (\mathfrak m) = \dim _{\kappa (\mathfrak m)} \mathfrak m/\mathfrak m^2.$

Proof. Consider the exact sequence

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

of Lemma 10.131.9. We would like to show that the first map is an isomorphism. Since $k$ is algebraically closed the composition $k \to \kappa (\mathfrak m)$ is an isomorphism by Theorem 10.34.1. So the surjection $S \to \kappa (\mathfrak m)$ splits as a map of $k$ -algebras, and Lemma 10.131.10 shows that the sequence above is exact on the left. Since $\Omega _{\kappa (\mathfrak m)/k} = 0$ , we win. $\square$

Comments (2)

Comment #1212 by Max on December 19, 2014 at 00:47

Hello! What do you think of the following proof of this lemma? It is somewhat quicker.

Consider an exact sequence
${\mathfrak m}/{\mathfrak m}^2 \to \Omega_{S/k} \otimes_S \kappa({\mathfrak m}) \to \Omega_{\kappa({\mathfrak m})/k} \to 0.$ We would like to show that the first map is an isomorphism. Since $k$ is algebraically closed the composition $k \to S \to \kappa({\mathfrak m})$ is an isomorphism. So the surjection $S \to \kappa({\mathfrak m})$ splits as a map of $k$ -algebras, and lemma 02HP shows that the sequence above is exact on the left. Since $\Omega_{\kappa({\mathfrak m})} = 0$ , we win.

Comment #1243 by Johan on January 06, 2015 at 20:53

Wonderful! See change here.

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