Exercise 5.12

Question

Let $G$ be a finite group of automorphisms of a ring $A$, and let $A^G$ denote the subring of $G$-invariants, that is of all $x \in A$ such that $\sigma(x) = x$ for all $\sigma \in G$. Prove that $A$ is integral over $A^G$.

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$]
  %^$A^G$ is a subring since if $x,y \in A^G$, $xy = \sigma(x)\sigma(y) = \sigma(xy)$, $x\pm y = \sigma(x)\pm\sigma(y) = \sigma(x\pm y)$, and $1 \in A^G$ by the fact that $0 
e x = \sigma(x) = \sigma(1x) = \sigma(1)x \implies \sigma(1) = 1$.
  Let $x \in A$ and consider
  \begin{equation*}
    f(t) = \prod_{\sigma \in G} (t - \sigma(x)) = (t-x)\prod_{\sigma 
e e}(t-\sigma(x)),
  \end{equation*}
  since $e \in G$. Thus, $f(x) = 0$, and since the highest order term of $f(t)$ is $t^{|G|}$, we see that $f$ is monic. We then show $f(t)$ has coefficients in $A^G$. We see that the coefficients of $f$ are elementary symmetric polynomials, i.e., of the form
  \begin{equation*}
    a_k = \sum_{i_1 < \cdots < i_k} \sigma_{i_1}(x)\cdots\sigma_{i_k}(x),
  \end{equation*}
  where we enumerate $G=\{\sigma_i\}$. $	au(a_k) = a_k$ for any $	au\in G$, for $\sigma$ just permutes terms in the summation; thus, $a_k \in A^G$ for all $k$, and $A$ is integral over $A^G$.
\end{proof}

Exercise 5.12

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Exercise 5.12

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