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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 42730–42742 (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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