Lemma 29.54.9. Let $X$ be a scheme with locally finitely many irreducible components. The normalization morphism $\nu : X^\nu \to X$ is an isomorphism if and only if $X$ is normal.
Proof. If $\nu $ is an isomorphism, then $X$ is normal by Lemma 29.54.5. Conversely, suppose $X$ is normal. By Lemma 29.53.6 and Properties, Lemma 28.7.5 we may assume $X$ is integral. By Lemma 29.54.5 the morphism $\nu $ is integral and $X^\nu $ has a unique irreducible component (so it is an integral scheme). We conclude from Lemmas 29.54.7 and 29.54.8. $\square$
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