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:

• $R_k:A_{𝔭}$ is a regular local ring for any prime ideal $𝔭$ of height ≤ k.
• $S_k:\mathrm{depth}{A}_{𝔭}\ge \inf \{k,\mathrm{ht}(𝔭)\}$ for any prime ideal $𝔭$.

The statement is:

• A is a reduced ring $\iff R_0,S_1$ hold.
• A is a normal ring $\iff R_1,S_2$ hold.
• A is a Cohen–Macaulay ring $\iff S_k$ 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 ${𝔭}_i,1\le i\le r$ are the minimal prime ideals of A, then the total ring of fractions K of A is the direct product of the residue fields $\kappa ({𝔭}_i)=Q(A/{𝔭}_i)$: 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 $\kappa ({𝔭}_i)$ 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

$(f/g{)}^n+a_1(f/g{)}^{n-1}+\dots +a_n=0$

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 $𝔭$ of $A/gA$ has height one. This is because if $𝔭$ has height greater than one, then $𝔭$ would contain a non zero divisor in $A/gA$. However, $𝔭$ is associated to the zero ideal in $A/gA$ so it can only contain zero divisors, see here. By the condition R₁, the localization $A_{𝔭}$ is integrally closed and so $\varphi (f)\in \varphi (g)A_{𝔭}$, where $\varphi :A\to A_{𝔭}$ is the localization map, since the integral equation persists after localization. If $gA={\cap }_i{𝔮}_i$ is the primary decomposition, then, for any i, the radical of ${𝔮}_i$ is an associated prime $𝔭$ of $A/gA$ and so $f\in {\varphi }^{-1}({𝔮}_i{A}_{𝔭})={𝔮}_i$; the equality here is because ${𝔮}_i$ is a $𝔭$-primary ideal. Hence, the assertion holds.

Necessity

Suppose A is a normal ring. For S₂, let $𝔭$ 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 $𝔭$. By definition, there is an element g in A such that $𝔭=\{x\in A|xg\equiv 0\text{ mod }fA\}$ and $g\in ̸fA$. Put y = g/f in the total ring of fractions. If $y𝔭\subset 𝔭$, then $𝔭$ 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, $y𝔭=A$ or $𝔭=f/gA$, which implies $𝔭$ has height one (Krull's principal ideal theorem).

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

References

• Template:EGA
• H. Matsumura, Commutative algebra, 1970.