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: The statement is: Items 1, 3 trivially follow from the definitions. Item 2 is much deeper.
 * $$R_k: A_{\mathfrak{p}}$$ is a regular local ring for any prime ideal $$\mathfrak{p}$$ of height ≤ k.
 * $$S_k: \operatorname{depth} A_{\mathfrak{p}} \ge \inf \{k, \operatorname{ht}(\mathfrak{p}) \}$$ for any prime ideal $$\mathfrak{p}$$.
 * 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.

For an integral domain, the criterion is due to Krull. The general case is due to Serre.

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 Aei'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, ai'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 $$\phi(f) \in \phi(g)A_{\mathfrak{p}}$$, 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 $$\mathfrak{p} = f/g A$$, 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. $$\square$$