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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!

user237522
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  • What about $s_1=s_2=x$ and $k=y$ ? – Carot Nov 02 '19 at 18:17
  • @Carot, thank you for your nice comment. Actually, I had in mind algebraically independent $s_1,s_2$. I apologize for not mentioning this. – user237522 Nov 02 '19 at 18:19
  • Ok then what about $k=y$ and $s_1=x$ and $s_2=y^4$ ? – Carot Nov 03 '19 at 11:09
  • @Carot, nice, thank you very much. Please, do you think that there exists an additional mild condition that guarantees a positive answer to my question? – user237522 Nov 03 '19 at 14:57
  • Well as long as I can take $k=y$, no. And I don't see a very convincing condition excluding this (and similar) case(s). – Carot Nov 03 '19 at 16:05
  • @Carot, I see, thank you. – user237522 Nov 03 '19 at 16:08
  • @Carot, please, do you think that the following additional condition guarantees that $\mathbb{C}(s_1,s_2)=\mathbb{C}(x,y^2)$? The additional condition is: $y \notin \mathbb{C}(s_i,k)$, $1 \leq i \leq 2$. – user237522 Nov 20 '19 at 18:02

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