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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: