Questions on ring fixed by automorphisms

Question

Suppose $A$ is an integral domain and $F$ is its field of fractions. Let $G \leq Aut(F)$ be a group of automorphisms of $F$. Assume $g(a) \subseteq A$ for all $g \in G$ and let $A^G$ be the fixed ring $$A^G=\{a \in A\ \mid g(a)=a, \forall g\in G\}.$$ I have managed to show that $A$ is integral over $A^G$ by considering the polynomial $\prod_{g\in G}(x-g(a))$ and showing it has coefficients in $A^G[x]$. Clearly it is monic, so the result follows. I have some more questions about this setup. I would also like to show that the fraction field of $A^G$ is $F^G$ and I am not sure how to do it. I think I might have to use the norm or trace maps $N,T:A \rightarrow A^G$ $$N(a)=\prod_{g\in G} g(a),$$ $$T(a)=\sum_{g\in G} g(a).$$ Also, I would like to show that $A$ Noetherian $\implies A^G$ Noetherian and $A$ Dedekind $\implies A^G$ Dedekind.

Answer 1

It is true that the fraction field of $A^G$ is $F^G$, at least when $G$ is finite (which you seem to assume is your proof of integrality).

It is well-known that that the trace map of a separable extension is always surjective. Hence (the extension $F/F^G$ is Galois, so in particular separable) any element of $F^G$ is of the form $Tr(\alpha)=Tr(\frac{x}{y})$ for $\alpha\in F$, $x,y\in A$. This is equal to $$\sum_{g\in G} \frac{gx}{gy}$$ $$\frac{\sum_{g\in G} \frac{N(y)}{gy}gx}{N(y)}$$ Both the numerator and denominator are in $A^G$, which shows that $F^G \subseteq \text{Frac}(A^G)$ (the reverse inclusion is trivial).

For Noetherianness, my idea is that Eakin-Nagata theorem might be helpful; we would need to show that $A$ is finitely generated as an $A^G$ module. I can't see right now how to do that, though.

Answer 2

There is the following lemma: If $A \subseteq B$ is a ring extension with $x_1,..,x_k \in B$ with $B\cong A[x_1,..,x_k]$ and such that each $x_i$ is integral over $A$ then $B$ is a finitely generated $A$-module.

It seems the following holds:

Proposition. Assume $G$ is a finite group. If there is a finite set of elements $x_1,..,x_k \in A$ with $A\cong A^G[x_1,..,x_k]$ (as a ring) then $A$ is a finitely generated $A^G$-module. Hence if $A$ is Noetherian and $A$ is a finitely generated $A^G$-algebra it follows $A^G$ is Noetherian.