Lemma 28.7.5. Let $X$ be a scheme such that any quasi-compact open has a finite number of irreducible components. The following are equivalent:

1. $X$ is normal, and

2. $X$ is a disjoint union of normal integral schemes.

Proof. It is immediate from the definitions that (2) implies (1). Let $X$ be a normal scheme such that every quasi-compact open has a finite number of irreducible components. If $X$ is affine then $X$ satisfies (2) by Algebra, Lemma 10.37.16. For a general $X$, let $X = \bigcup X_ i$ be an affine open covering. Note that also each $X_ i$ has but a finite number of irreducible components, and the lemma holds for each $X_ i$. Let $T \subset X$ be an irreducible component. By the affine case each intersection $T \cap X_ i$ is open in $X_ i$ and an integral normal scheme. Hence $T \subset X$ is open, and an integral normal scheme. This proves that $X$ is the disjoint union of its irreducible components, which are integral normal schemes. $\square$

## Comments (2)

Comment #3046 by Dario Weißmann on

Last sentence of the proof: "There are only finitely many by assumption." We only assume that for quasi-compact opens. But the proof was already finished, so we are fine without that last sentence.

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• 4 comment(s) on Section 28.7: Normal schemes

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