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:

$R_{k}:A_{{{\mathfrak {p}}}}$ is a regular local ring for any prime ideal ${\mathfrak {p}}$ of height ≤ k.
$S_{k}:\operatorname {depth}A_{{{\mathfrak {p}}}}\geq \inf\{k,\operatorname {ht}({\mathfrak {p}})\}$ for any prime ideal ${\mathfrak {p}}$ .^[1]

The statement is:

A is a reduced ring $\Leftrightarrow R_{0},S_{1}$ hold.
A is a normal ring $\Leftrightarrow R_{1},S_{2}$ hold.
A is a Cohen–Macaulay ring $\Leftrightarrow S_{k}$ 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 IV. Theorem 5.8.6.)

Suppose A satisfies S₂ and R₁. Then A in particular satisfies S₁ and R₀; hence, it is reduced. If ${\mathfrak {p}}_{i},\,1\leq i\leq r$ are the minimal prime ideals of A, then the total ring of fractions K of A is the direct product of the residue fields $\kappa ({\mathfrak {p}}_{i})=Q(A/{\mathfrak {p}}_{i})$ : see total ring of fractions of a reduced ring. That means we can write $1=e_{1}+\dots +e_{r}$ where $e_{i}$ are idempotents in $\kappa ({\mathfrak {p}}_{i})$ and such that $e_{i}e_{j}=0,\,i\neq j$ . 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}}+\dots +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 ${\mathfrak {p}}$ of $A/gA$ has height one. By the condition R₁, the localization $A_{{{\mathfrak {p}}}}$ is integrally closed and so $\phi (f)\in \phi (g)A_{{{\mathfrak {p}}}}$ , where $\phi :A\to A_{{{\mathfrak {p}}}}$ is the localization map, since the integral equation persists after localization. If $gA=\cap _{i}{\mathfrak {q}}_{i}$ is the primary decomposition, then, for any i, the radical of ${\mathfrak {q}}_{i}$ is an associated prime ${\mathfrak {p}}$ of $A/gA$ and so $f\in \phi ^{{-1}}({\mathfrak {q}}_{i}A_{{{\mathfrak {p}}}})={\mathfrak {q}}_{i}$ ; the equality here is because ${\mathfrak {q}}_{i}$ is a ${\mathfrak {p}}$ -primary ideal. Hence, the assertion holds.

Necessity

Suppose A is a normal ring. For S₂, let ${\mathfrak {p}}$ 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 ${\mathfrak {p}}$ . By definition, there is an element g in A such that ${\mathfrak {p}}=\{x\in A|xg\equiv 0{\text{ mod }}fA\}$ and $g\not \in fA$ . Put y = g/f in the total ring of fractions. If $y{\mathfrak {p}}\subset {\mathfrak {p}}$ , then ${\mathfrak {p}}$ 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, $y{\mathfrak {p}}=A$ or ${\mathfrak {p}}=f/gA$ , which implies ${\mathfrak {p}}$ has height one (Krull's principal ideal theorem).

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

Notes

↑ EGA IV, § 5.7.

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. MR 0199181. doi:10.1007/bf02684322.
H. Matsumura, Commutative algebra, 1970.

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