Serre's criterion for normality explained

In algebra, Serre's criterion for normality, introduced by Jean-Pierre Serre, gives necessary and sufficient conditions for a commutative Noetherian ring A to be a normal ring. The criterion involves the following two conditions for A:

R _{k:} A _{ak{p} } is a regular local ring for any prime ideal ak{p} of height ≤ k.S _{k:} \operatorname{depth} A _{ak{p} } \ge \inf \ for any prime ideal ak{p} .The statement is:\Leftrightarrow R _{0,} S _{1} hold.

• A is a normal ring

\Leftrightarrow R _{1,} S _{2} hold.\Leftrightarrow S _{k} hold for all k.Items 1, 3 trivially follow from the definitions. Item 2 is much deeper.

For an integral domain, the criterion is due to Krull. The general case is due to Serre.

Proof

Sufficiency

(After EGA IV₂. Theorem 5.8.6.)

Suppose A satisfies S₂ and R₁. Then A in particular satisfies S₁ and R₀; hence, it is reduced. If

ak{p} _{i,} 1 \le i \le r are the minimal prime ideals of A, then the total ring of fractions K of A is the direct product of the residue fields \kappa( ak{p} _{i)} = Q(A/ ak{p} _{i)} : see total ring of fractions of a reduced ring. That means we can write 1 = e _{1} + ... + e _{r} where e _{i} are idempotents in \kappa( ak{p} _{i)} and such that e _{i} e _{j} = 0 , i \ne j . Now, if A is integrally closed in K, then each e _{i} is integral over A and so is in A; consequently, A is a direct product of integrally closed domains Ae_i's and we are done. Thus, it is enough to show that A is integrally closed in K.

For this end, suppose

(f/g) ^{n} + a _{1} (f/g) ^{n-1} + ... + a _{n} = 0 where all f, g, a_i's are in A and g is moreover a non-zerodivisor. We want to show:f \in gA .Now, the condition S₂ says that gA is unmixed of height one; i.e., each associated primes ak{p} of A/gA has height one. This is because if ak{p} has height greater than one, then ak{p} would contain a non zero divisor in A/gA . However, ak{p} is associated to the zero ideal in A/gA so it can only contain zero divisors, see here. By the condition R₁, the localization A _{ak{p} } is integrally closed and so \phi(f) \in \phi(g)A _{ak{p} }, where \phi: A \to A _{ak{p} } is the localization map, since the integral equation persists after localization. If gA = \cap _{i} ak{q} _{i} is the primary decomposition, then, for any i, the radical of ak{q} _{i} is an associated prime ak{p} of A/gA and so f \in \phi ^{-1} ( ak{q} _{i} A _{ak{p} }) = \mathfrak_i; the equality here is because ak{q} _{i} is a ak{p} -primary ideal. Hence, the assertion holds.

Necessity

Suppose A is a normal ring. For S₂, let

ak{p} be an associated prime of A/fA for a non-zerodivisor f; we need to show it has height one. Replacing A by a localization, we can assume A is a local ring with maximal ideal ak{p} . By definition, there is an element g in A such that ak{p} = \{ x \in A | xg \equiv 0 mod fA \} and g \not\in fA . Put y = g/f in the total ring of fractions. If y ak{p} \subset ak{p} , then ak{p} is a faithful A[y] -module and is a finitely generated A-module; consequently, y is integral over A and thus in A, a contradiction. Hence, y ak{p} = A or ak{p} = f/g A , which implies ak{p} has height one (Krull's principal ideal theorem).

For R₁, we argue in the same way: let

ak{p} be a prime ideal of height one. Localizing at ak{p} we assume ak{p} is a maximal ideal and the similar argument as above shows that ak{p} is in fact principal. Thus, A is a regular local ring. \square

References

• • H. Matsumura, Commutative algebra, 1970.