Lemma 10.155.4 (Serre's criterion for normality). Let $R$ be a Noetherian ring. The following are equivalent:

$R$ is a normal ring, and

$R$ has properties $(R_1)$ and $(S_2)$.

** Normal equals R1 plus S2. **

[IV, Theorem 5.8.6, EGA]

Lemma 10.155.4 (Serre's criterion for normality). Let $R$ be a Noetherian ring. The following are equivalent:

$R$ is a normal ring, and

$R$ has properties $(R_1)$ and $(S_2)$.

**Proof.**
Proof of (1) $\Rightarrow $ (2). Assume $R$ is normal, i.e., all localizations $R_{\mathfrak p}$ at primes are normal domains. In particular we see that $R$ has $(R_0)$ and $(S_1)$ by Lemma 10.155.3. Hence it suffices to show that a local Noetherian normal domain $R$ of dimension $d$ has depth $\geq \min (2, d)$ and is regular if $d = 1$. The assertion if $d = 1$ follows from Lemma 10.118.7.

Let $R$ be a local Noetherian normal domain with maximal ideal $\mathfrak m$ and dimension $d \geq 2$. Apply Lemma 10.118.2 to $R$. It is clear that $R$ does not fall into cases (1) or (2) of the lemma. Let $R \to R'$ as in (4) of the lemma. Since $R$ is a domain we have $R \subset R'$. Since $\mathfrak m$ is not an associated prime of $R'$ there exists an $x \in \mathfrak m$ which is a nonzerodivisor on $R'$. Then $R_ x = R'_ x$ so $R$ and $R'$ are domains with the same fraction field. But finiteness of $R \subset R'$ implies every element of $R'$ is integral over $R$ (Lemma 10.35.3) and we conclude that $R = R'$ as $R$ is normal. This means (4) does not happen. Thus we get the remaining possibility (3), i.e., $\text{depth}(R) \geq 2$ as desired.

Proof of (2) $\Rightarrow $ (1). Assume $R$ satisfies $(R_1)$ and $(S_2)$. By Lemma 10.155.3 we conclude that $R$ is reduced. Hence it suffices to show that if $R$ is a reduced local Noetherian ring of dimension $d$ satisfying $(S_2)$ and $(R_1)$ then $R$ is a normal domain. If $d = 0$, the result is clear. If $d = 1$, then the result follows from Lemma 10.118.7.

Let $R$ be a reduced local Noetherian ring with maximal ideal $\mathfrak m$ and dimension $d \geq 2$ which satisfies $(R_1)$ and $(S_2)$. By Lemma 10.36.16 it suffices to show that $R$ is integrally closed in its total ring of fractions $Q(R)$. Pick $x \in Q(R)$ which is integral over $R$. Then $R' = R[x]$ is a finite ring extension of $R$ (Lemma 10.35.5). Because $\dim (R_\mathfrak p) < d$ for every nonmaximal prime $\mathfrak p \subset R$ we have $R_\mathfrak p = R'_\mathfrak p$ by induction. Hence the support of $R'/R$ is $\{ \mathfrak m\} $. It follows that $R'/R$ is annihilated by a power of $\mathfrak m$ (Lemma 10.61.4). By Lemma 10.118.2 this contradicts the assumption that the depth of $R$ is $\geq 2 = \min (2, d)$ and the proof is complete. $\square$

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