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.
• $S_k:\mathrm{depth}{A}_{\mathfrak{𝔭}}\ge \inf \{k,\mathrm{ht}(\mathfrak{𝔭})\}$ for any prime ideal $\mathfrak{𝔭}$.

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 ${\mathfrak{𝔭}}_i,1\le i\le r$ are the minimal prime ideals of A, then the total ring of fractions K of A is the direct product of the residue fields $\kappa ({\mathfrak{𝔭}}_i)=Q(A/{\mathfrak{𝔭}}_i)$: 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 $e_i{e}_j=0,i\ne j$. 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 $\mathfrak{𝔭}$ of $A/gA$ has height one. This is because if $\mathfrak{𝔭}$ has height greater than one, then $\mathfrak{𝔭}$ would contain a non zero divisor in $A/gA$. However, $\mathfrak{𝔭}$ is associated to the zero ideal in $A/gA$ so it can only contain zero divisors, see here. By the condition R₁, the localization $A_{\mathfrak{𝔭}}$ is integrally closed and so $\varphi (f)\in \varphi (g)A_{\mathfrak{𝔭}}$, where $\varphi :A\to A_{\mathfrak{𝔭}}$ 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 $f\in {\varphi }^{-1}({\mathfrak{𝔮}}_i{A}_{\mathfrak{𝔭}})={\mathfrak{𝔮}}_i$; 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 $g\in ̸fA$. 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. $◻$

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.