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. $\square$

Comment #1212 by on

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

Consider an exact sequence 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.

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