## 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 42472–42489 (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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