Let $k=\mathbb C$, let $K$ be a finite extension of the field $k(X_1,X_2,X_3)$ of rational functions in $3$ variables, let $L/K$ be a finite Galois extension of commutative fields and let the Galois group of $L/K$ be $G$. I'm mostly interested in the case when $L/K$ is a biquadratic extension i.e. $G$ is isomorphic to $C_2^2:= \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$.
$\newcommand{\ord}{\operatorname{ord}}$ Let $D$ be a division algebra which is isomorphic to a crossed product $L^{\alpha}G$ for some cocycle $\alpha$. For $g\in G$ let $\ord(g)$ be the order of $g$. When $g\in G$ is considered as an element of $L^\alpha G$ then conjugation by $g^{\ord(g)}$ is a trivial automorphism of $L$, so by maximality of $L$ we have $g^{\ord(g)}\in L$.
Let $R(\alpha)$ be the smallest subfield of $L$ which contains $k$ and all the elements $g^{\ord(g)}$, $g\in G$.
Question What is $R(\alpha)$?
In some cases $R(\alpha)=L$, and generally speaking I'm interested in conditions which assure that $R(\alpha)=L$. Here is an example of a more concrete question.
Question Suppose that $D$ is as above and $R(\alpha)=L$. Suppose that $L_0$ is another maximal commutative subfield of $D$. Let $H$ be the Galois group of $L_0/K$. I think for the specific case of $G\cong C_2^2$ it follows that $H\cong C_2^2$,but to be on the safe side assume that this is the case. Note that by the Skolem-Noether theorem we have that $D$ is isomorphic to a crossed product $L_0^{\alpha_0} H$. Under what extra conditions can we find a cocycle $\alpha_0$ such that $D$ is isomorphic to $L_0^{\alpha_0} H$ and such that $R(\alpha_0) = L_0$ ?
I'm not sure if there is a definite answer, so any remarks would be highly appreciated.
