Lemma 10.138.1. 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

**Proof.**
Consider the exact sequence

of Lemma 10.130.9. 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 10.33.1. So the surjection $S \to \kappa (\mathfrak m)$ splits as a map of $k$-algebras, and Lemma 10.130.10 shows that the sequence above is exact on the left. Since $\Omega _{\kappa (\mathfrak m)/k} = 0$, we win. $\square$

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