The Stacks Project


Tag 0333

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

Lemma 10.155.15. Let $R$ be a Noetherian domain. Assume

  1. there exists a nonzero $f \in R$ such that $R_f$ is normal, and
  2. for every maximal ideal $\mathfrak m \subset R$ the local ring $R_{\mathfrak m}$ is N-1.

Then $R$ is N-1.

Proof. Let $K$ be the fraction field of $R$. Suppose that $R \subset R' \subset K$ is a finite extension of $R$ contained in $K$. Note that $R_f = R'_f$ since $R_f$ is already normal. Hence by Lemma 10.155.14 the set of primes $\mathfrak p' \in \mathop{\rm Spec}(R')$ with $R'_{\mathfrak p'}$ non-normal is closed in $\mathop{\rm Spec}(R')$. Since $\mathop{\rm Spec}(R') \to \mathop{\rm Spec}(R)$ is closed the image of this set is closed in $\mathop{\rm Spec}(R)$. For such a ring $R'$ denote $Z_{R'} \subset \mathop{\rm Spec}(R)$ this image.

Pick a maximal ideal $\mathfrak m \subset R$. Let $R_{\mathfrak m} \subset R_{\mathfrak m}'$ be the integral closure of the local ring in $K$. By assumption this is a finite ring extension. By Lemma 10.35.11 we can find finitely many elements $r_1, \ldots, r_n \in K$ integral over $R$ such that $R_{\mathfrak m}'$ is generated by $r_1, \ldots, r_n$ over $R_{\mathfrak m}$. Let $R' = R[x_1, \ldots, x_n] \subset K$. With this choice it is clear that $\mathfrak m \not \in Z_{R'}$.

As $\mathop{\rm Spec}(R)$ is quasi-compact, the above shows that we can find a finite collection $R \subset R'_i \subset K$ such that $\bigcap Z_{R'_i} = \emptyset$. Let $R'$ be the subring of $K$ generated by all of these. It is finite over $R$. Also $Z_{R'} = \emptyset$. Namely, every prime $\mathfrak p'$ lies over a prime $\mathfrak p'_i$ such that $(R'_i)_{\mathfrak p'_i}$ is normal. This implies that $R'_{\mathfrak p'} = (R'_i)_{\mathfrak p'_i}$ is normal too. Hence $R'$ is normal, in other words $R'$ is the integral closure of $R$ in $K$. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 42909–42919 (see updates for more information).

    \begin{lemma}
    \label{lemma-characterize-N-1}
    Let $R$ be a Noetherian domain.
    Assume
    \begin{enumerate}
    \item there exists a nonzero $f \in R$ such that $R_f$ is normal, and
    \item for every maximal ideal $\mathfrak m \subset R$
    the local ring $R_{\mathfrak m}$ is N-1.
    \end{enumerate}
    Then $R$ is N-1.
    \end{lemma}
    
    \begin{proof}
    Let $K$ be the fraction field of $R$.
    Suppose that $R \subset R' \subset K$ is a finite
    extension of $R$ contained in $K$. Note that $R_f = R'_f$ since
    $R_f$ is already normal. Hence by Lemma \ref{lemma-openness-normal-locus}
    the set of primes
    $\mathfrak p' \in \Spec(R')$ with $R'_{\mathfrak p'}$ non-normal
    is closed in $\Spec(R')$. Since $\Spec(R') \to \Spec(R)$
    is closed the image of this set is closed in $\Spec(R)$.
    For such a ring $R'$ denote $Z_{R'} \subset \Spec(R)$ this image.
    
    \medskip\noindent
    Pick a maximal ideal $\mathfrak m \subset R$.
    Let $R_{\mathfrak m} \subset R_{\mathfrak m}'$ be the integral
    closure of the local ring in $K$. By assumption this is
    a finite ring extension. By Lemma \ref{lemma-integral-closure-localize}
    we can find finitely
    many elements $r_1, \ldots, r_n \in K$ integral over $R$ such that
    $R_{\mathfrak m}'$ is generated by $r_1, \ldots, r_n$ over $R_{\mathfrak m}$.
    Let $R' = R[x_1, \ldots, x_n] \subset K$. With this choice it is clear
    that $\mathfrak m \not \in Z_{R'}$.
    
    \medskip\noindent
    As $\Spec(R)$ is quasi-compact, the above shows that we can
    find a finite collection $R \subset R'_i \subset K$ such that
    $\bigcap Z_{R'_i} = \emptyset$. Let $R'$ be the subring of $K$
    generated by all of these. It is finite over $R$. Also $Z_{R'} = \emptyset$.
    Namely, every prime $\mathfrak p'$ lies over a prime $\mathfrak p'_i$
    such that $(R'_i)_{\mathfrak p'_i}$ is normal. This implies
    that $R'_{\mathfrak p'} = (R'_i)_{\mathfrak p'_i}$ is normal too.
    Hence $R'$ is normal, in other words
    $R'$ is the integral closure of $R$ in $K$.
    \end{proof}

    Comments (0)

    There are no comments yet for this tag.

    There are also 3 comments on Section 10.155: Commutative Algebra.

    Add a comment on tag 0333

    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?