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.
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.
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:
The scheme X is normal.
For every affine open U \subset X the ring \mathcal{O}_ X(U) is normal.
There exists an affine open covering X = \bigcup U_ i such that each \mathcal{O}_ X(U_ i) is normal.
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
Lemma 28.7.3. A normal scheme is reduced.
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:
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 10.37.16. 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:
X is normal, and
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:
X is normal, and
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!
Lemma 28.7.9.slogan 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
Comment #2576 by Johan on
Comment #8816 by Hsueh-Yung Lin on
Comment #9273 by Stacks project on