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28.7 Normal schemes

Recall that a ring $R$ is said to be normal if all its local rings are normal domains, see Algebra, Definition 10.37.11. A normal domain is a domain which is integrally closed in its field of fractions, see Algebra, Definition 10.37.1. Thus it makes sense to define a normal scheme as follows.

Definition 28.7.1. A scheme $X$ is normal if and only if for all $x \in X$ the local ring $\mathcal{O}_{X, x}$ is a normal domain.

This seems to be the definition used in EGA, see [0, 4.1.4, EGA]. Suppose $X = \mathop{\mathrm{Spec}}(A)$, and $A$ is reduced. Then saying that $X$ is normal is not equivalent to saying that $A$ is integrally closed in its total ring of fractions. However, if $A$ is Noetherian then this is the case (see Algebra, Lemma 10.37.16).

Lemma 28.7.2. Let $X$ be a scheme. The following are equivalent:

  1. The scheme $X$ is normal.

  2. For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is normal.

  3. There exists an affine open covering $X = \bigcup U_ i$ such that each $\mathcal{O}_ X(U_ i)$ is normal.

  4. There exists an open covering $X = \bigcup X_ j$ such that each open subscheme $X_ j$ is normal.

Moreover, if $X$ is normal then every open subscheme is normal.

Proof. This is clear from the definitions. $\square$

Proof. Immediate from the definitions. $\square$

Lemma 28.7.4. Let $X$ be an integral scheme. Then $X$ is normal if and only if for every nonempty affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is a normal domain.

Proof. This follows from Algebra, Lemma 10.37.10. $\square$

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. $\square$

Lemma 28.7.6. Let $X$ be a Noetherian scheme. The following are equivalent:

  1. $X$ is normal, and

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

Proof. This is a special case of Lemma 28.7.5 because a Noetherian scheme has a Noetherian underlying topological space (Lemma 28.5.5 and Topology, Lemma 5.9.2). $\square$

Lemma 28.7.7. Let $X$ be a locally Noetherian scheme. The following are equivalent:

  1. $X$ is normal, and

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

Proof. Omitted. Hint: This is purely topological from Lemma 28.7.6. $\square$

Remark 28.7.8. Let $X$ be a normal scheme. If $X$ is locally Noetherian then we see that $X$ is integral if and only if $X$ is connected, see Lemma 28.7.7. But there exists a connected affine scheme $X$ such that $\mathcal{O}_{X, x}$ is a domain for all $x \in X$, but $X$ is not irreducible, see Examples, Section 110.6. This example is even a normal scheme (proof omitted), so beware!

slogan

Lemma 28.7.9. Let $X$ be an integral normal scheme. Then $\Gamma (X, \mathcal{O}_ X)$ is a normal domain.

Proof. Set $R = \Gamma (X, \mathcal{O}_ X)$. It is clear that $R$ is a domain. Suppose $f = a/b$ is an element of its fraction field which is integral over $R$. Say we have $f^ d + \sum _{i = 0, \ldots , d - 1} a_ i f^ i = 0$ with $a_ i \in R$. Let $U \subset X$ be a nonempty affine open. Since $b \in R$ is not zero and since $X$ is integral we see that also $b|_ U \in \mathcal{O}_ X(U)$ is not zero. Hence $a/b$ is an element of the fraction field of $\mathcal{O}_ X(U)$ which is integral over $\mathcal{O}_ X(U)$ (because we can use the same polynomial $f^ d + \sum _{i = 0, \ldots , d - 1} a_ i|_ U f^ i = 0$ on $U$). Since $\mathcal{O}_ X(U)$ is a normal domain (Lemma 28.7.4), we see that $f_ U = (a|_ U)/(b|_ U) \in \mathcal{O}_ X(U)$. It is clear that $f_ U|_ V = f_ V$ whenever $V \subset U \subset X$ are nonempty affine open. Hence the local sections $f_ U$ glue to an element $g \in R = \Gamma (X, \mathcal{O}_ X)$. Then $bg$ and $a$ restrict to the same element of $\mathcal{O}_ X(U)$ for all $U$ as above, hence $bg = a$, in other words, $g$ maps to $f$ in the fraction field of $R$. $\square$

Comments (4)

Comment #2543 by Grayson Jorgenson on

There appears to be a minor typo in the proof of Lemma 27.7.9: the summations are indexed , but should be indexed or the exponents changed from to .

Comment #8816 by Hsueh-Yung Lin on

Small typo: In the proof of Lemma 28.7.9, I think we should cite Lemma 28.7.4 instead of Lemma 28.7.2.

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