# 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.

Lemma 1.If A is normal, then for all , we have

**Note**

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*.

**Proof**

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 . ♦

Lemma 2.Suppose for all , , we have . Then

,

where intersection occurs in .

**Proof**

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 . ♦

**Exercise A**

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*.

**Note**

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, .

**Note**

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.

**Proof**

(⇒) 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*. ♦

# Hartog’s Theorem

Definition.If V is an affine k-variety, we say V is

normalif is a normal domain.If is a non-empty closed subset, corresponding to radical ideal , the

codimensionof 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

.

**Note**

Thus, if a rational function *f* on *V* is not regular, its “set of irregularity” has codimension 1.

**Proof**

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.

Corollary 1.If are as in proposition 1, then is not an affine variety.

**Proof**

The inclusion induces an isomorphism . If *V*–*W* were affine, *f* would also be an isomorphism, which is absurd since *f* is not surjective. ♦

# Dimensionality

Finally we wish to look at the dimension of a general quasi-projective variety; first we have a general definition.

Definition.Let X be a non-empty topological space; we consider all chains

of irreducible closed subsets of X. The

lengthof the chain is k; theKrull dimensionof X (denoted ) is the supremum of the lengths of all such chains.The

dimensionof a quasi-projective variety is its Krull dimension.

Note that when , the Krull dimension of *X* is the Krull dimension of *A*.

Lemma 3.If Y is a non-empty subspace of X then .

**Proof**

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.

Proposition 2.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.

**Proof**

First, is an irreducible subset of *X* by proposition 2 here. Next note that is a non-empty open subset of *C*; hence by proposition 1 here is irreducible. It remains to show:

.

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:

Corollary 2.Let U and X be as above; if C’ is an irreducible component of U, then is an irreducible component of X.

**Proof**

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.

Lemma 4.If is an open cover for an irreducible space X, and each , then .

**Proof**

(≥) 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:

Proposition 3.Let V be a quasi-projective variety. If V is irreducible and W is any non-empty open subset of V, then .

In particular, .

**Proof**

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. ♦