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Tag 0350

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

Example 10.155.2. Let $k$ be a field. The domain $R = k[x_1, x_2, x_3, \ldots]$ is N-2, but not Noetherian. The reason is the following. Suppose that $R \subset L$ and the field $L$ is a finite extension of the fraction field of $R$. Then there exists an integer $n$ such that $L$ comes from a finite extension $k(x_1, \ldots, x_n) \subset L_0$ by adjoining the (transcendental) elements $x_{n + 1}, x_{n + 2}$, etc. Let $S_0$ be the integral closure of $k[x_1, \ldots, x_n]$ in $L_0$. By Proposition 10.156.16 below it is true that $S_0$ is finite over $k[x_1, \ldots, x_n]$. Moreover, the integral closure of $R$ in $L$ is $S = S_0[x_{n + 1}, x_{n + 2}, \ldots]$ (use Lemma 10.36.8) and hence finite over $R$. The same argument works for $R = \mathbf{Z}[x_1, x_2, x_3, \ldots]$.

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 42581–42598 (see updates for more information).

    \begin{example}
    \label{example-Japanese-not-Noetherian}
    Let $k$ be a field. The domain $R = k[x_1, x_2, x_3, \ldots]$ is N-2,
    but not Noetherian. The reason is the following. Suppose that $R \subset L$
    and the field $L$ is a finite extension of the fraction field of $R$.
    Then there exists an integer $n$ such that $L$ comes from a finite
    extension $k(x_1, \ldots, x_n) \subset L_0$ by adjoining
    the (transcendental) elements $x_{n + 1}, x_{n + 2}$, etc.
    Let $S_0$ be the integral
    closure of $k[x_1, \ldots, x_n]$ in $L_0$. By
    Proposition \ref{proposition-ubiquity-nagata} below
    it is true that $S_0$ is finite over $k[x_1, \ldots, x_n]$.
    Moreover, the integral closure of $R$ in $L$ is
    $S = S_0[x_{n + 1}, x_{n + 2}, \ldots]$ (use
    Lemma \ref{lemma-polynomial-domain-normal}) and
    hence finite over $R$. The same argument works for
    $R = \mathbf{Z}[x_1, x_2, x_3, \ldots]$.
    \end{example}

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