Let $\beta$ be the following involution on $\mathbb{C}[x,y]$ (= an automorphism of order two): $\beta: (x,y) \mapsto (x,-y)$.
Let $s_1,s_2 \in S_{\beta}$ and $k \in K_{\beta}$.
Assume that the following conditions are satisfied:
(1) $\mathbb{C}(s_1,s_1,k)=\mathbb{C}(x,y)$.
(2) $s_1,s_2$ are algebraically independent over $\mathbb{C}$.
My revised question:
Could one find an additional ('weak') condition that will guarantee that $\mathbb{C}(s_1,s_2)=\mathbb{C}(x,y^2)$?
Please notice that my original question was: "Is it true that $\mathbb{C}(s_1,s_2)=\mathbb{C}(x,y^2)$?" (See older edits).
An attempt to answer: Take $s_1=x, s_2=y^4, k=xy$. Perhaps a plausible additional condition that will guarantee that $\mathbb{C}(s_1,s_2)=\mathbb{C}(x,y^2)$ is: $y \notin \mathbb{C}(s_i,k)$, $1 \leq i \leq 2$.
I have asked the above question at MO.
Thank you very much!