27.7 Normal schemes

Recall that a ring $R$ is said to be normal if all its local rings are normal domains, see Algebra, Definition 10.36.11. A normal domain is a domain which is integrally closed in its field of fractions, see Algebra, Definition 10.36.1. Thus it makes sense to define a normal scheme as follows.

Definition 27.7.1. 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 10.36.16).

Lemma 27.7.2. 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 27.7.4. Let $X$ be an integral scheme. Then $X$ is normal if and only if for every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is a normal domain.

Proof. This follows from Algebra, Lemma 10.36.10. $\square$

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:

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 10.36.16. 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. $\square$

Lemma 27.7.6. 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 27.7.5 because a Noetherian scheme has a Noetherian underlying topological space (Lemma 27.5.5 and Topology, Lemma 5.9.2. $\square$

Lemma 27.7.7. 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 27.7.6. $\square$

Remark 27.7.8. 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 27.7.7. 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 104.5. This example is even a normal scheme (proof omitted), so beware!

Lemma 27.7.9. 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 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 27.7.2), we see that $f_ U = (a|_ U)/(b|_ U) \in \mathcal{O}_ X(U)$. It is easy to see that $f_ U|_ V = f_ V$ whenever $V \subset U \subset X$ are affine open. Hence the local sections $f_ U$ glue to a global section $f$ as desired. $\square$

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 $i = 1,...,d$, but should be indexed $i = 0,...,d-1$ or the exponents changed from $f^i$ to $f^{d - i}$.

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