I would like to ask if the following is true:
Suppose we have an integral domain $A$ and a group action $G$ on $A$. We consider $A^{G}$ the subring of $A$ fixed by $G$. Let $L$ be the field of fractions of $A^{G}$ and $K$ the field of fractions of $A.$ Can we safely say that $K$ is a Galois extension of $L$ and if so, can we say that $A$ is a free $A^{G}$ module?
Thanks!
Partial answer only: Let $K$ be a field and $G$ be a finite group. Then the extension $K/K^G$ is Galois.
Proof: for every $x\in K$ the orbit $Gx$ of $x$ under $G$ is invariant under $G$. The coefficients of the separable polynomial
$ f_x:=\prod\limits_{a\in Gx} (X-a) $
thus are in $K^G$. One concludes that $K/K^G$ is separable algebraic.
Now let $g\in K^G[X]$ be an irreducible polynomial having $x\in K$ as a root. Then all elements of $Gx$ are roots of $g$. Hence $f_x$ divides $g$, that is $g=f_x$, which shows that $g$ splits completely in $K$. Consequently $K/K^G$ is normal.
The argument also shows that the ring extension $A/A^G$ is integral.
The assertion is wrong if $G$ is infinite: let $k$ be an infinite field and let $G$ be the group of all $k$-automorphisms $g$ of the rational function field $K:=k(X)$ with the property $g(X)=\frac{aX+b}{cX+d}$. Then $k(X)^G=k$.
A rather simple case in which freeness holds: let $A$ be a 1-dimensional, noetherian, integrally closed domain and assume that $A^G$ is local. Then $A/A^G$ is free.
Proof: $A^G$ is 1-dimensional; by the Lying-over-Theorem there exists a maximal ideal $M$ of $A$ lying over the maximal ideal of $A^G$. By assumption $A_M$ is a discrete valuation ring.
One always has $A\cap K^G = A^G$. Consequently $A_M\cap K^G=A^G$, hence $A^G$ is a discrete valuation ring.
Moreover $A^G$ is integrally closed, that is $A$ is the integral closure of $A^G$ in the Galois extension $K/K^G$, thus $A/A^G$ is finite.
Finitely generated, torsion-free modules over a discrete valution ring are free.
Note that one can check locality of $A^G$ in $A$ itself: $G$ acts on the maximal ideals of $A$, and $A^G$ is local iff all maximal ideals lie in the same orbit.
In general in my opinion the problem of freeness seems to be difficult: several things can go wrong. If $A/A^G$ is free it must have rank equal to $|G|$ -- in particular the ring extension must be finite. It is difficult to ensure this without assumptions about noetherianity of $A^G$. But it seems that $A$ noetherian does not imply that $A^G$ is noetherian. And even in the noetherian and finite case, moreover assuming that $A$ is integrally closed, the extension $A/A^G$ is usually not free.