$\mathbb{C}(s_1,s_2,k)=\mathbb{C}(x,y)$, where $s_1,s_2$ are symmetric and $k$ is skew-symmetric

42 Views Asked by At

Let $\beta: \mathbb{C}[x,y] \to \mathbb{C}[x,y]$ be the involution (= $\mathbb{C}$-algebra automorphism of $\mathbb{C}[x,y]$ of degree two) defined by $(x,y) \mapsto (x,-y)$.

It is not difficult to see that a symmetric element of $\mathbb{C}[x,y]$ w.r.t. $\beta$ is of the following form: $a_{2n}y^{2n}+a_{2n-2}y^{2n-2}+\cdots+a_2y^2+a_0$, where $a_{2i} \in \mathbb{C}[x]$, and a skew-symmetric element of $\mathbb{C}[x,y]$ w.r.t. $\beta$ is of the following form: $b_{2m+1}y^{2m+1}+b_{2m-1}y^{2m-1}+\cdots+b_3y^3+b_1y$, where $b_{2j+1} \in \mathbb{C}[x]$.

Denote the set of symmetric elements of $\mathbb{C}[x,y]$ w.r.t. $\beta$ by $S_{\beta}$ and the set of skew-symmetric elements of $\mathbb{C}[x,y]$ w.r.t. $\beta$ by $K_{\beta}$.

Let $s_1,s_2 \in S_{\beta}$ and $k \in K_{\beta}$.

Assume that:

(i) $\{s_1,s_2\}$, $\{s_1,k\}$, $\{s_2,k\}$ are algebraically independent over $\mathbb{C}$. (The three $\{s_1,s_2,k\}$ are of course algebraically dependent over $\mathbb{C}$).

(ii) $\mathbb{C}(s_1,s_2,k)=\mathbb{C}(x,y)$.

(iii) $\mathbb{C}(s_1,k) \subsetneq \mathbb{C}(x,y)$, $\mathbb{C}(s_2,k) \subsetneq \mathbb{C}(x,y)$ and $\mathbb{C}(s_1,s_2) \subsetneq \mathbb{C}(x,y^2)$.

Example: $s_1=x^2+y^2$, $s_2=x^3y^2$, $k=x^3y^3$.

Question: Is it possible to find a 'specific' form of such $s_1,s_2,k$?

Edit: What happens in the special case $k=y$? I am not able to find an example; if we take $s_1=x^2, s_2=xy^2$, then $\mathbb{C}(s_2,k)=\mathbb{C}(x,y)$, so condition (iii) is not satisfied.

See also this similar (actually equivalent) question; notice that here if we write $k=s_3y$, where $s_3 \in S_{\beta}$, we get that $\mathbb{C}(x,y)= \mathbb{C}(s_1,s_2,k)= \mathbb{C}(s_1,s_2,s_3y) \subseteq \mathbb{C}(s_1,s_2,s_3,y)$, hence $\mathbb{C}(x,y)=\mathbb{C}(s_1,s_2,s_3,y)$, which is assumed in the linked question.

Remark: A description of symmetric and skew-symmetric elements of $\mathbb{C}(x,y)$ w.r.t. $\beta$ can be found here.

Thank you very much!