## Tag `0357`

Chapter 27: Properties of Schemes > Section 27.7: Normal schemes

Lemma 27.7.5. Let $X$ be a scheme such that any quasi-compact open has a finite number of irreducible components. The following are equivalent:

- $X$ is normal, and
- $X$ 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 10.36.16. 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. There are only finitely many by assumption. $\square$

The code snippet corresponding to this tag is a part of the file `properties.tex` and is located in lines 813–821 (see updates for more information).

```
\begin{lemma}
\label{lemma-normal-locally-finite-nr-irreducibles}
Let $X$ be a scheme such that any quasi-compact open has a finite number
of irreducible components. The following are equivalent:
\begin{enumerate}
\item $X$ is normal, and
\item $X$ is a disjoint union of normal integral schemes.
\end{enumerate}
\end{lemma}
\begin{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 \ref{algebra-lemma-characterize-reduced-ring-normal}.
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. There are only finitely many
by assumption.
\end{proof}
```

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