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Tag 032I

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

Lemma 10.155.5. Let $R$ be a domain. Let $R \subset S$ be a quasi-finite extension of domains (for example finite). Assume $R$ is N-2 and Noetherian. Then $S$ is N-2.

Proof. Let $K = f.f.(R) \subset L = f.f.(S)$. Note that this is a finite field extension (for example by Lemma 10.121.2 (2) applied to the fibre $S \otimes_R K$, and the definition of a quasi-finite ring map). Let $S'$ be the integral closure of $R$ in $S$. Then $S'$ is contained in the integral closure of $R$ in $L$ which is finite over $R$ by assumption. As $R$ is Noetherian this implies $S'$ is finite over $R$. By Lemma 10.122.15 there exist elements $g_1, \ldots, g_n \in S'$ such that $S'_{g_i} \cong S_{g_i}$ and such that $g_1, \ldots, g_n$ generate the unit ideal in $S$. Hence it suffices to show that $S'$ is N-2 by Lemmas 10.155.3 and 10.155.4. Thus we have reduced to the case where $S$ is finite over $R$.

Assume $R \subset S$ with hypotheses as in the lemma and moreover that $S$ is finite over $R$. Let $M$ be a finite field extension of the fraction field of $S$. Then $M$ is also a finite field extension of $f.f(R)$ and we conclude that the integral closure $T$ of $R$ in $M$ is finite over $R$. By Lemma 10.35.16 we see that $T$ is also the integral closure of $S$ in $M$ and we win by Lemma 10.35.15. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 42520–42527 (see updates for more information).

    \begin{lemma}
    \label{lemma-quasi-finite-over-Noetherian-japanese}
    Let $R$ be a domain.
    Let $R \subset S$ be a quasi-finite extension of domains
    (for example finite).
    Assume $R$ is N-2 and Noetherian.
    Then $S$ is N-2.
    \end{lemma}
    
    \begin{proof}
    Let $K = f.f.(R) \subset L = f.f.(S)$. Note that this is
    a finite field extension (for example by
    Lemma \ref{lemma-isolated-point-fibre} (2)
    applied to the fibre $S \otimes_R K$, and the definition of a
    quasi-finite ring map).
    Let $S'$ be the integral closure of $R$ in $S$.
    Then $S'$ is contained in the integral closure of $R$ in $L$
    which is finite over $R$ by assumption. As $R$ is Noetherian this
    implies $S'$ is finite over $R$.
    By Lemma \ref{lemma-quasi-finite-open-integral-closure}
    there exist elements $g_1, \ldots, g_n \in S'$
    such that $S'_{g_i} \cong S_{g_i}$ and such that $g_1, \ldots, g_n$
    generate the unit ideal in $S$. Hence it suffices to show that
    $S'$ is N-2 by Lemmas \ref{lemma-localize-N} and \ref{lemma-Japanese-local}.
    Thus we have reduced to the case where $S$ is finite over $R$.
    
    \medskip\noindent
    Assume $R \subset S$ with hypotheses as in the lemma and moreover
    that $S$ is finite over $R$. Let $M$ be a finite field extension
    of the fraction field of $S$. Then $M$ is also a finite field extension
    of $f.f(R)$ and we conclude that the integral closure $T$ of $R$ in
    $M$ is finite over $R$. By Lemma \ref{lemma-integral-closure-transitive}
    we see that $T$ is also the integral closure of $S$ in $M$ and we win by
    Lemma \ref{lemma-integral-permanence}.
    \end{proof}

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