For $x\in A$, consider
$$
f_x(T)=\prod_{\sigma\in G}(T-\sigma(x)).
$$
This is monic, and every $\tau\in G$ permutes its factors, so all coefficients lie in $A^G$. Since $f_x(x)=0$, every $x\in A$ is integral over $A^G$. Thus
$$
\boxed{A\text{ is integral over }A^G.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 4 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-101/4/a/ii/solution