## Tag `03B7`

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

Example 10.155.9. Lemma 10.155.8 does not work if the ring is not Noetherian. For example consider the action of $G = \{+1, -1\}$ on $A = \mathbf{C}[x_1, x_2, x_3, \ldots]$ where $-1$ acts by mapping $x_i$ to $-x_i$. The invariant ring $R = A^G$ is the $\mathbf{C}$-algebra generated by all $x_ix_j$. Hence $R \subset A$ is not finite. But $R$ is a normal domain with fraction field $K = L^G$ the $G$-invariants in the fraction field $L$ of $A$. And clearly $A$ is the integral closure of $R$ in $L$.

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 42699–42711 (see updates for more information).

```
\begin{example}
\label{example-bad-invariants}
Lemma \ref{lemma-Noetherian-normal-domain-finite-separable-extension}
does not work if the ring is not Noetherian.
For example consider the action of $G = \{+1, -1\}$ on
$A = \mathbf{C}[x_1, x_2, x_3, \ldots]$ where $-1$ acts by
mapping $x_i$ to $-x_i$. The invariant ring $R = A^G$ is
the $\mathbf{C}$-algebra generated by all $x_ix_j$. Hence
$R \subset A$ is not finite. But $R$ is a normal domain
with fraction field $K = L^G$ the $G$-invariants in the fraction field
$L$ of $A$. And clearly $A$ is the integral closure of $R$ in
$L$.
\end{example}
```

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