## Tag `031S`

Chapter 10: Commutative Algebra > Section 10.151: Serre's criterion for normality

**Normal equals R1 plus S2.**

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. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 41383–41394 (see updates for more information).

```
\begin{lemma}[Serre's criterion for normality]
\label{lemma-criterion-normal}
\begin{slogan}
Normal equals R1 plus S2.
\end{slogan}
Let $R$ be a Noetherian ring.
The following are equivalent:
\begin{enumerate}
\item $R$ is a normal ring, and
\item $R$ has properties $(R_1)$ and $(S_2)$.
\end{enumerate}
\end{lemma}
\begin{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 \ref{lemma-criterion-reduced}. 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 \ref{lemma-characterize-dvr}.
\medskip\noindent
Let $R$ be a local Noetherian normal domain with maximal ideal
$\mathfrak m$ and dimension $d \geq 2$. Apply
Lemma \ref{lemma-hart-serre-loc-thm} 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 \ref{lemma-finite-is-integral})
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.
\medskip\noindent
Proof of (2) $\Rightarrow$ (1). Assume $R$ satisfies $(R_1)$ and $(S_2)$.
By Lemma \ref{lemma-criterion-reduced} 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 \ref{lemma-characterize-dvr}.
\medskip\noindent
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 \ref{lemma-characterize-reduced-ring-normal}
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 \ref{lemma-characterize-finite-in-terms-of-integral}).
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 \ref{lemma-Noetherian-power-ideal-kills-module}).
By Lemma \ref{lemma-hart-serre-loc-thm} this
contradicts the assumption that the depth of $R$ is $\geq 2 = \min(2, d)$
and the proof is complete.
\end{proof}
```

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