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:

  • is a regular local ring for any prime ideal of height ≤ k.
  • for any prime ideal .[1]

The statement is:

  • A is a reduced ring hold.
  • A is a normal ring hold.
  • A is a Cohen–Macaulay ring 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 are the minimal prime ideals of A, then the total ring of fractions K of A is the direct product of the residue fields : see total ring of fractions of a reduced ring. That means we can write where are idempotents in and such that . Now, if A is integrally closed in K, then each is integral over A and so is in A; consequently, A is a direct product of integrally closed domains Aei's and we are done. Thus, it is enough to show that A is integrally closed in K.

For this end, suppose

where all f, g, ai's are in A and g is moreover a non-zerodivisor. We want to show:

.

Now, the condition S2 says that is unmixed of height one; i.e., each associated primes of has height one. This is because if has height greater than one, then would contain a non zero divisor in . However, is associated to the zero ideal in so it can only contain zero divisors, see here. By the condition R1, the localization is integrally closed and so , where is the localization map, since the integral equation persists after localization. If is the primary decomposition, then, for any i, the radical of is an associated prime of and so ; the equality here is because is a -primary ideal. Hence, the assertion holds.

Necessity

Suppose A is a normal ring. For S2, let be an associated prime of 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 . By definition, there is an element g in A such that and . Put y = g/f in the total ring of fractions. If , then is a faithful -module and is a finitely generated A-module; consequently, is integral over A and thus in A, a contradiction. Hence, or , which implies has height one (Krull's principal ideal theorem).

For R1, we argue in the same way: let be a prime ideal of height one. Localizing at we assume is a maximal ideal and the similar argument as above shows that is in fact principal. Thus, A is a regular local ring.

Notes

References

  • Grothendieck, Alexandre; Dieudonné, Jean (1965). "Éléments de géométrie algébrique: IV. Étude locale des schémas et des morphismes de schémas, Seconde partie". Publications Mathématiques de l'IHÉS. 24. doi:10.1007/bf02684322. MR 0199181.
  • H. Matsumura, Commutative algebra, 1970.
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