Lemma 10.151.4 (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 10.151.4 (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 10.151.3. 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 10.151.3 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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (3)

Comment #847 by Jason Starr on

Comment #2729 by Ariyan Javanpeykar on

Comment #2854 by Johan on

There are also: