Serre’s Criterion for Normality
Throughout this article, fix an algebraically closed field k.
In this section, A denotes a noetherian domain. We will describe Serre’s criterion, which is a necessary and sufficient condtion for A to be normal. In the following section, we will relate the results here to an interesting example from the last article.
If A is normal, then for all , we have
Since A is a domain, it has only one minimal prime: 0. Hence all associated primes of A/aA have height at least 1. Lemma 1 thus says principal ideals of a normal domain have no embedded primes.
We need to show has height 1. Pick such that has annihilator , i.e. . Since bA is finitely generated, we localize both sides at to obtain , the unique maximal ideal of . Thus
and . We claim that ; if not, and by the adjugate matrix trick (see proof of proposition 6 here), is integral over B. This contradicts the fact that B is normal. Hence is an invertible ideal so B is a dvr, and . ♦
Suppose for all , , we have . Then
where intersection occurs in .
Let , where , ; we need to show . Write for its primary decomposition with associated primes all of height 1. For each i we have . But since all are minimal in . Thus
by proposition 2 here and . ♦
Let for an irreducible affine variety V, and be a prime ideal with corresponding subvariety . Prove that is the set of all rational functions on V which are regular at some point of W.
We already know holds for all domains; geometrically, this means if (for irreducible V) is regular at each point, then f can be represented by the same polynomial globally. The condition in lemma 2 is notably stronger; geometrically, it says if f is regular on an open dense subset of every codimension 1 subvariety, then it is regular everywhere.
Theorem (Serre’s Criterion).
A noetherian domain is normal if and only if the following conditions both hold.
- All , for , have height 1.
- For each of height 1, is a dvr.
When that happens, .
In the context of algebraic geometry, the first condition says “subvarieties cut out by a single equation have no embedded components” while the second says “the set of singular points has codimension at least 2” (this will be elaborated in later articles). Thus normality can fail in two different ways: hidden (embedded) components or too many singular points.
(⇒) Condition 1 follows from lemma 1; condition 2 follows from proposition 6 here.
(⇐) By lemma 2, condition 1 gives us , the final statement. By condition 2, each is normal; hence so is A. ♦
If V is an affine k-variety, we say V is normal if is a normal domain.
If is a non-empty closed subset, corresponding to radical ideal , the codimension of W in V is .
As a consequence of the above result, we have:
Proposition 1 (Hartog’s Theorem).
Let V be an affine normal k-variety. If is a non-empty closed subset of codimension at least 2, then the inclusion induces an isomorphism of coordinate rings
Thus, if a rational function f on V is not regular, its “set of irregularity” has codimension 1.
Suppose , considered as a rational function on V. For each of height 1, f is regular on , a dense open subset of . Hence by exercise A, and by Serre’s criterion, we have
This gives the following generalization of example 7 from the previous article.
If are as in proposition 1, then is not an affine variety.
The inclusion induces an isomorphism . If V–W were affine, f would also be an isomorphism, which is absurd since f is not surjective. ♦
Finally we wish to look at the dimension of a general quasi-projective variety; first we have a general definition.
Let X be a non-empty topological space; we consider all chains
of irreducible closed subsets of X. The length of the chain is k; the Krull dimension of X (denoted ) is the supremum of the lengths of all such chains.
The dimension of a quasi-projective variety is its Krull dimension.
Note that when , the Krull dimension of X is the Krull dimension of A.
If Y is a non-empty subspace of X then .
Let be a chain of irreducible closed subsets of Y. Taking their closures in X we have
By proposition 2 here, each is a closed irreducible subset of X. Furthermore by a general result in point-set topology (see proposition 3 here), the closure of any subset in Y is . Hence so for any i. Thus we get a chain of irreducible closed subsets of X of length k. ♦
To proceed, we need the following correspondence.
Let U be a non-empty open subset of a topological space X. Then there is a bijective correspondence:
where is the closure of C’ in X.
For the first statement, is a non-empty open subset of C so it is dense in C. The second statement holds for any closed subset C’ of U as noted above, so we are done. ♦
In particular, we have:
Let U and X be as above; if C’ is an irreducible component of U, then is an irreducible component of X.
By the correspondence in proposition 2, since C’ is a maximal irreducible closed subset of U, cannot be properly contained in any irreducible closed subset of X. ♦
Let us return to results on dimensionality.
If is an open cover for an irreducible space X, and each , then .
(≥) holds by lemma 3. For (≤), suppose is a chain of closed irreducible subsets of X. Pick an i such that . By proposition 2 we get a chain of irreducible closed subsets of of length k :
So . ♦
Finally, we have:
Let V be a quasi-projective variety. If V is irreducible and W is any non-empty open subset of V, then .
In particular, .
If V is affine, this follows from proposition 2 here. For the general case, V is an open subset of some projective V’ so it suffices to assume V is projective, i.e. a closed subset of some .
By lemma 4, for some we have . Since is affine, by what we just showed . On the other hand by lemma 3, so equality holds throughout. ♦