Serre's criterion for normality

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:

• [math]\displaystyle{ R_k: A_{\mathfrak{p}} }[/math] is a regular local ring for any prime ideal [math]\displaystyle{ \mathfrak{p} }[/math] of height ≤ k.
• [math]\displaystyle{ S_k: \operatorname{depth} A_{\mathfrak{p}} \ge \inf \{k, \operatorname{ht}(\mathfrak{p}) \} }[/math] for any prime ideal [math]\displaystyle{ \mathfrak{p} }[/math].

The statement is:

• A is a reduced ring [math]\displaystyle{ \Leftrightarrow R_0, S_1 }[/math] hold.
• A is a normal ring [math]\displaystyle{ \Leftrightarrow R_1, S_2 }[/math] hold.
• A is a Cohen–Macaulay ring [math]\displaystyle{ \Leftrightarrow S_k }[/math] 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 IV. Theorem 5.8.6.)

Suppose A satisfies S₂ and R₁. Then A in particular satisfies S₁ and R₀; hence, it is reduced. If [math]\displaystyle{ \mathfrak{p}_i, \, 1 \le i \le r }[/math] are the minimal prime ideals of A, then the total ring of fractions K of A is the direct product of the residue fields [math]\displaystyle{ \kappa(\mathfrak{p}_i) = Q(A/\mathfrak{p}_i) }[/math]: see total ring of fractions of a reduced ring. That means we can write [math]\displaystyle{ 1 = e_1 + \dots + e_r }[/math] where [math]\displaystyle{ e_i }[/math] are idempotents in [math]\displaystyle{ \kappa(\mathfrak{p}_i) }[/math] and such that [math]\displaystyle{ e_i e_j = 0, \, i \ne j }[/math]. Now, if A is integrally closed in K, then each [math]\displaystyle{ e_i }[/math] 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

[math]\displaystyle{ (f/g)^n + a_1 (f/g)^{n-1} + \dots + a_n = 0 }[/math]

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

[math]\displaystyle{ f \in gA }[/math].

Now, the condition S₂ says that [math]\displaystyle{ gA }[/math] is unmixed of height one; i.e., each associated primes [math]\displaystyle{ \mathfrak{p} }[/math] of [math]\displaystyle{ A/gA }[/math] has height one. By the condition R₁, the localization [math]\displaystyle{ A_{\mathfrak{p}} }[/math] is integrally closed and so [math]\displaystyle{ \phi(f) \in \phi(g)A_{\mathfrak{p}} }[/math], where [math]\displaystyle{ \phi: A \to A_{\mathfrak{p}} }[/math] is the localization map, since the integral equation persists after localization. If [math]\displaystyle{ gA = \cap_i \mathfrak{q}_i }[/math] is the primary decomposition, then, for any i, the radical of [math]\displaystyle{ \mathfrak{q}_i }[/math] is an associated prime [math]\displaystyle{ \mathfrak{p} }[/math] of [math]\displaystyle{ A/gA }[/math] and so [math]\displaystyle{ f \in \phi^{-1}(\mathfrak{q}_i A_{\mathfrak{p}}) = \mathfrak{q}_i }[/math]; the equality here is because [math]\displaystyle{ \mathfrak{q}_i }[/math] is a [math]\displaystyle{ \mathfrak{p} }[/math]-primary ideal. Hence, the assertion holds.

Necessity

Suppose A is a normal ring. For S₂, let [math]\displaystyle{ \mathfrak{p} }[/math] be an associated prime of [math]\displaystyle{ A/fA }[/math] 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 [math]\displaystyle{ \mathfrak{p} }[/math]. By definition, there is an element g in A such that [math]\displaystyle{ \mathfrak{p} = \{ x \in A | xg \equiv 0 \text{ mod }fA \} }[/math] and [math]\displaystyle{ g \not\in fA }[/math]. Put y = g/f in the total ring of fractions. If [math]\displaystyle{ y \mathfrak{p} \subset \mathfrak{p} }[/math], then [math]\displaystyle{ \mathfrak{p} }[/math] is a faithful [math]\displaystyle{ A[y] }[/math]-module and is a finitely generated A-module; consequently, [math]\displaystyle{ y }[/math] is integral over A and thus in A, a contradiction. Hence, [math]\displaystyle{ y \mathfrak{p} = A }[/math] or [math]\displaystyle{ \mathfrak{p} = f/g A }[/math], which implies [math]\displaystyle{ \mathfrak{p} }[/math] has height one (Krull's principal ideal theorem).

For R₁, we argue in the same way: let [math]\displaystyle{ \mathfrak{p} }[/math] be a prime ideal of height one. Localizing at [math]\displaystyle{ \mathfrak{p} }[/math] we assume [math]\displaystyle{ \mathfrak{p} }[/math] is a maximal ideal and the similar argument as above shows that [math]\displaystyle{ \mathfrak{p} }[/math] is in fact principal. Thus, A is a regular local ring. [math]\displaystyle{ \square }[/math]

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. http://www.numdam.org/articles/PMIHES_1965__24__5_0. 
• H. Matsumura, Commutative algebra, 1970.