Lemma 29.54.12. Let X be an irreducible, geometrically unibranch scheme. The normalization morphism \nu : X^\nu \to X is a universal homeomorphism.
Proof. We have to show that \nu is integral, universally injective, and surjective, see Lemma 29.45.5. By Lemma 29.54.5 the morphism \nu is integral. Let x \in X and set A = \mathcal{O}_{X, x}. Since X is irreducible we see that A has a single minimal prime \mathfrak p and A_{red} = A/\mathfrak p. By Lemma 29.54.4 the stalk A' = (\nu _*\mathcal{O}_{X^\nu })_ x is the integral closure of A in the fraction field of A_{red}. By More on Algebra, Definition 15.106.1 we see that A' has a single prime \mathfrak m' lying over \mathfrak m_ x \subset A and \kappa (\mathfrak m')/\kappa (x) is purely inseparable. Hence \nu is bijective (hence surjective) and universally injective by Lemma 29.10.2. \square
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