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Tag 032H

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

Lemma 10.155.4. Let $R$ be a domain. Let $f_1, \ldots, f_n \in R$ generate the unit ideal. If each domain $R_{f_i}$ is N-1 then so is $R$. Same for N-2.

Proof. Assume $R_{f_i}$ is N-2 (or N-1). Let $L$ be a finite extension of the fraction field of $R$ (equal to the fraction field in the N-1 case). Let $S$ be the integral closure of $R$ in $L$. By Lemma 10.35.11 we see that $S_{f_i}$ is the integral closure of $R_{f_i}$ in $L$. Hence $S_{f_i}$ is finite over $R_{f_i}$ by assumption. Thus $S$ is finite over $R$ by Lemma 10.23.2. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 42503–42508 (see updates for more information).

    \begin{lemma}
    \label{lemma-Japanese-local}
    Let $R$ be a domain. Let $f_1, \ldots, f_n \in R$ generate the
    unit ideal. If each domain $R_{f_i}$ is N-1 then so is $R$.
    Same for N-2.
    \end{lemma}
    
    \begin{proof}
    Assume $R_{f_i}$ is N-2 (or N-1).
    Let $L$ be a finite extension of the fraction field of $R$ (equal to
    the fraction field in the N-1 case). Let $S$ be the integral
    closure of $R$ in $L$. By Lemma \ref{lemma-integral-closure-localize}
    we see that $S_{f_i}$ is the integral closure of $R_{f_i}$ in $L$.
    Hence $S_{f_i}$ is finite over $R_{f_i}$ by assumption.
    Thus $S$ is finite over $R$ by Lemma \ref{lemma-cover}.
    \end{proof}

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