# 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 $a \in A - \{0\}$, we have

$\mathfrak p \in \mathrm{Ass}_A (A/aA) \implies \mathrm{ht} \mathfrak p = 1.$

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 $\mathfrak p \in \mathrm{Ass}_A (A/aA)$ has height 1. Pick $b\in A$ such that $b+aA \in A/aA$ has annihilator $\mathrm{Ann}_A (b+aA) = \mathfrak p$, i.e. $(aA : bA) = \mathfrak p$. Since bA is finitely generated, we localize both sides at $\mathfrak p$ to obtain $(aA_{\mathfrak p} : bA_{\mathfrak p}) = \mathfrak m$, the unique maximal ideal of $B := A_{\mathfrak p}$. Thus

$b\mathfrak m \subseteq aB \implies \overbrace{(ba^{-1})}^{\in \mathrm{Frac} A}\mathfrak m \subseteq B$

and $ba^{-1} \not\in B$. We claim that $ba^{-1} \mathfrak m = B$; if not, $ba^{-1} \mathfrak m \subseteq \mathfrak m$ and by the adjugate matrix trick (see proof of proposition 6 here), $ba^{-1}$ is integral over B. This contradicts the fact that B is normal. Hence $\mathfrak m$ is an invertible ideal so B is a dvr, and $\mathrm{ht} \mathfrak p = \dim B = 1$. ♦

Lemma 2.

Suppose for all $\mathfrak p \in \mathrm{Ass}_A (A/aA)$, $a\in A-\{0\}$, we have $\mathrm{ht} \mathfrak p =1$. Then

$A = \bigcap_{\mathrm{ht} \mathfrak p = 1} A_{\mathfrak p}$,

where intersection occurs in $\mathrm{Frac} A$.

Proof

Let $\frac a b \in \text{RHS}$, where $a, b \in A$, $b\ne 0$; we need to show $a \in bA$. Write $bA = \mathfrak q_1 \cap \ldots \cap \mathfrak q_n$ for its primary decomposition with associated primes $\mathfrak p_i = r(\mathfrak q_i)$ all of height 1. For each i we have $\frac a b \in A_{\mathfrak p_i} \implies a \in bA_{\mathfrak p_i}$. But $bA_{\mathfrak p_i} = \mathfrak q_i A_{\mathfrak p_i}$ since all $\mathfrak p_i$ are minimal in $V(aA)$. Thus

$a \in \mathfrak q_i A_{\mathfrak p_i} \cap A = \mathfrak q_i$

by proposition 2 here and $a \in \cap_i \mathfrak q_i = bA$. ♦

Exercise A

Let $A = k[V]$ for an irreducible affine variety V, and $\mathfrak p \subset A$ be a prime ideal with corresponding subvariety $W = V(\mathfrak p) \subset V$. Prove that $A_{\mathfrak p}$ is the set of all rational functions on V which are regular at some point of W.

Note

We already know $A = \cap_{\mathfrak m \text{ max.}} A_{\mathfrak m}$ holds for all domains; geometrically, this means if $f:V \to \mathbb A^1$ (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 $A$ is normal if and only if the following conditions both hold.

1. All $\mathfrak p \in \mathrm{Ass}_A (A/aA)$, for $a\in A-\{0\}$, have height 1.
2. For each $\mathfrak p$ of height 1, $A_{\mathfrak p}$ is a dvr.

When that happens, $A = \cap_{\mathrm{ht} \mathfrak p = 1} A_{\mathfrak p}$.

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 $A = \cap_{\mathrm{ht} \mathfrak p = 1} A_{\mathfrak p}$, the final statement. By condition 2, each $A_{\mathfrak p}$ is normal; hence so is A. ♦

# Hartog’s Theorem

Definition.

If V is an affine k-variety, we say V is normal if $k[V]$ is a normal domain.

If $W\subseteq V$ is a non-empty closed subset, corresponding to radical ideal $\mathfrak a \subsetneq k[V]$, the codimension of W in V is $\mathrm{ht} \mathfrak a$.

As a consequence of the above result, we have:

Proposition 1 (Hartog’s Theorem).

Let V be an affine normal k-variety. If $W\subsetneq V$ is a non-empty closed subset of codimension at least 2, then the inclusion $V-W \subseteq V$ induces an isomorphism of coordinate rings

$k[V-W] \cong k[V]$.

Note

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

Proof

Suppose $f \in k[V-W]$, considered as a rational function on V. For each $\mathfrak p \subset k[V]$ of height 1, f is regular on $(V-W) \cap V(\mathfrak p) \ne \emptyset$, a dense open subset of $V(\mathfrak p)$. Hence by exercise A, $f\in A_{\mathfrak p}$ and by Serre’s criterion, we have

$f\in \bigcap_{\mathrm{ht} \mathfrak p = 1} A_{\mathfrak p} = A$. ♦

This gives the following generalization of example 7 from the previous article.

Corollary 1.

If $V, W$ are as in proposition 1, then $V-W$ is not an affine variety.

Proof

The inclusion $f : V-W \hookrightarrow V$ induces an isomorphism $f^* : k[V] \to k[V-W]$. If VW 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

$\emptyset \ne Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_k \subseteq X$

of irreducible closed subsets of X. The length of the chain is k; the Krull dimension of X (denoted $\dim X$) is the supremum of the lengths of all such chains.

The dimension of a quasi-projective variety is its Krull dimension.

Note that when $X = \mathrm{Spec} A$, the Krull dimension of X is the Krull dimension of A.

Lemma 3.

If Y is a non-empty subspace of X then $\dim Y \le \dim X$.

Proof

Let $Z_0 \subsetneq \ldots \subsetneq Z_k$ be a chain of irreducible closed subsets of Y. Taking their closures in X we have

$\emptyset\ne\overline Z_0 \subseteq \overline Z_1 \subseteq \ldots \subseteq \overline Z_k \subseteq X$.

By proposition 2 here, each $\overline Z_i$ 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 $Z\subseteq Y$ in Y is $\overline Z \cap Y$. Hence $\overline Z_i \cap Y = Z_i$ so $\overline Z_i \ne \overline Z_{i+1}$ 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 $\overline{C'}$ is the closure of C’ in X.

Proof

First, $\overline{C'}$ is an irreducible subset of X by proposition 2 here. Next note that $C\cap U$ is a non-empty open subset of C; hence by proposition 1 here $C\cap U$ is irreducible. It remains to show:

$\overline{C \cap U} = C, \quad \overline {C'} \cap U = C'$.

For the first statement, $C\cap U$ 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 $\overline {C'}$ is an irreducible component of X.

Proof

By the correspondence in proposition 2, since C’ is a maximal irreducible closed subset of U, $\overline{C'}$ cannot be properly contained in any irreducible closed subset of X. ♦

Let us return to results on dimensionality.

Lemma 4.

If $(U_i)$ is an open cover for an irreducible space X, and each $U_i \ne \emptyset$, then $\dim X = \sup \dim U_i$.

Proof

(≥) holds by lemma 3. For (≤), suppose $Z_0 \subsetneq \ldots \subsetneq Z_k$ is a chain of closed irreducible subsets of X. Pick an i such that $U_i \cap Z_0 \ne\emptyset$. By proposition 2 we get a chain of irreducible closed subsets of $U_i$ of length k :

$\emptyset \ne U_i \cap Z_0 \subseteq U_i \cap Z_1 \subseteq \ldots \subseteq U_i \cap Z_k$.

So $\dim U_i \ge k$. ♦

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 $\dim W = \dim V$.

In particular, $\dim \mathbb P^n_k = \dim \mathbb A^n_k = n$.

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 $\mathbb P^n$.

By lemma 4, for some $0\le i \le n$ we have $\dim V = \dim (U_i \cap V)$. Since $U_i \cap V$ is affine, by what we just showed $\dim (U_i \cap V) = \dim (W\cap U_i\cap V)$. On the other hand $\dim (W\cap U_i \cap V) \le \dim W \le \dim V$ by lemma 3, so equality holds throughout. ♦

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