This question asks for a help in proving the following exercise:
If $G$ is a finite group that acts on ring $A$, and $A^G$ is the subring consisting of elements of $A$ which are invariant under all $g\in G$, then $A$ is integral over $A^G$. (Atiyah and Macdonald, Chapter 5, Exercise 12).
I am wondering when the following claim may hold:
If $G$ is a finite group that acts on ring $A$, and $A^G$ is the subring consisting of elements of $A$ which are fixed under all $g\in G$, $A^G=\{a \in A|\forall g \in G, g(a)=a\}$, then $A$ is integral over $A^G$.
It seems that there are counter-examples to my above claim, hence I ask for an additional condition which will guarantee that it is true.
(What I had in mind: $A=k[x,y]$, $G=\{1,\alpha\}$, where $\alpha: (x,y) \mapsto (y,x)$. Then $A^G$ consists of the symmetric elements with respect to $\alpha$. I do not know if $A^G \subset A$ is integral or not).
Thank you very much for any help!
Edit: After reading the comments, I added to this question my exact definition of "fixed": $A^G=\{a \in A|\forall g \in G, g(a)=a\}$.