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Let $\beta$ be the following involution (=automorphism of order two) on $\mathbb{C}[x,y]$: $\beta: (x,y) \mapsto (x,-y)$. Denote the set of symmetric elements w.r.t. $\beta$ by $S_{\beta}$, and the set of skew-symmetric elements w.r.t. $\beta$ by $K_{\beta}$.

It is not difficult to see that if $s \in S_{\beta}$, then $s= a_{2n}y^{2n}+a_{2n-2}y^{2n-2}+\cdots+a_2y^2+a_0$, where $a_{2n},a_{2n-2},\ldots,a_2,a_0 \in \mathbb{C}[x]$, while if $k \in K_{\beta}$, then $a_{2n+1}y^{2n+1}+a_{2n-1}y^{2n-1}+\cdots+a_3y^3+a_1y$, where $a_{2n+1},a_{2n-1},\ldots,a_3,a_1 \in \mathbb{C}[x]$.

Let $s_1,s_2 \in S_{\beta}=\mathbb{C}[x,y^2]$ satisfy: (1) Each of $s_1,s_2$ is of (total) degree $\geq 1$. (2) $s_1,s_2$ are algebraically independent over $\mathbb{C}$. (3) $\mathbb{C}(s_1,s_2)=\mathbb{C}(x,y^2)$.

For example: $s_1=x, s_2=xy^2$.

Is it possible to find a general form of such a pair $s_1,s_2$?

Remarks: (i) If I am not wrong, if we replace condition $(3)$ to $\mathbb{C}[s_1,s_2]=\mathbb{C}[x,y^2]$, then the answer is as follows: Define $h: (x,y) \mapsto (x,y^2)$. Then a general form for such a pair $s_1,s_2$ is $s_1=(hg)(x), s_2=(hg)(y)$ where $g$ is an arbitrary automorphism of $\mathbb{C}[x,y]$.

(ii) Denote the group of automorphisms of $\mathbb{C}(x,y)$, the Cremona group, by $C$. If my argument in $(1)$ is correct, then perhaps an analog argument will solve my question as follows: Define $h: (x,y) \mapsto (x,y^2)$ ($h$ is extended to the field of fractions). Then a general form for $s_1,s_2$ is $s_1=(hg)(x), s_2=(hg)(y)$ where $g$ is an appropriate automorphism of $\mathbb{C}(x,y)$.

(iii) If my argument in $(2)$ is correct, then I would like to know how to describe $g \in C$; I think that an answer can be found in one of the references in this question, but I am not sure if I fully understand it.

(iv) Notice that not all elements of $C$ will be relevant to my question, since $s_1,s_2$ are assumed to be polynomials; for example, $C \ni g: (x,y) \mapsto (\frac 1x ,y)$ would yield $s_1=\frac 1x \notin \mathbb{C}[x,y]$ and $s_2=y^2$. (Or perhaps we should allow other options for $h$, for example, $h: (x,y) \mapsto (x, \frac{1}{y^2})$ and then we can take arbitrary $g \in C$).

Thank you very much!

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  • $\begingroup$ Why does $\beta$ play a role? Aren't you asking for $s_1,s_2\in\mathbb{C}[x,u]$, satisfying 1),2) and 3) with $y^2$ replaced by $u$? $\endgroup$ – Mohan May 9 '18 at 16:07
  • $\begingroup$ I guess you are right; do you know an answer to my question? The reason I mentioned $\beta$ is because I am also interested in the same question with $\alpha: (x,y) \mapsto (y,x)$ instead of $\beta$, but I wanted first to concentrate on $\beta$ hoping it is slightly easier to deal with than $\alpha$. $\endgroup$ – user237522 May 9 '18 at 16:49
  • $\begingroup$ I am not aware of any simple answer to it. Birational generators are notoriously difficult to find. There might be some answer using resultants (of $s_1-a, s_2-b$ for general $a,b\in\mathbb{C}$). $\endgroup$ – Mohan May 9 '18 at 17:31
  • $\begingroup$ Thank you for your comments. Please, could you elaborate a little your idea concerning resultants? $\endgroup$ – user237522 May 9 '18 at 18:04
  • $\begingroup$ $s_1,s_2$ form birational generators if and only if for general $a,b\in\mathbb{C}$, $s_1-a,s_2-b$ have exactly one common zero. Resultants help you calculate the common zeroes of $s_1-a, s_2-b$, so may be something can be done. $\endgroup$ – Mohan May 9 '18 at 18:20

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