The Stacks Project


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:

  1. $R$ is a normal ring, and
  2. $R$ 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 41486–41500 (see updates for more information).

    \begin{lemma}[Serre's criterion for normality]
    \label{lemma-criterion-normal}
    \begin{reference}
    \cite[IV, Theorem 5.8.6]{EGA}
    \end{reference}
    \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}

    References

    [EGA, IV, Theorem 5.8.6]

    Comments (3)

    Comment #847 by Jason Starr (site) on July 24, 2014 a 8:20 pm UTC

    Suggested slogan: Normal equals R1 plus S2.

    Comment #2729 by Ariyan Javanpeykar on August 1, 2017 a 11:54 pm UTC

    A reference for Serre's criterion: EGA IV2 Theorem 5.8.6

    Comment #2854 by Johan (site) on October 4, 2017 a 3:42 pm UTC

    Thanks added the reference here.

    There are also 2 comments on Section 10.151: Commutative Algebra.

    Add a comment on tag 031S

    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 lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?