Let $s_1,s_2, w \in \mathbb{C}[x,y]-\mathbb{C}$. Let $\alpha$ be the following involution, $\alpha: (x,y) \mapsto (y,x)$.
Assume that:
(1) Each of $s_1,s_2$ is symmetric w.r.t. $\alpha$.
(2) $w$ is not symmetric w.r.t. $\alpha$.
(3) $\mathbb{C}(s_1,s_2,w)=\mathbb{C}(x,y)$. (The notation is of fields of fractions).
I think that the following claim is true, but not sure how to prove it; any help in proving it is welcome!
Claim: (At least) one of the following happens:
- (At least) one of $\{s_1,s_2\}$ belongs to $\{\lambda(x+y),\mu xy\}_{\lambda,\mu \in \mathbb{C}}$.
- $w=\lambda y+as_1+bs_2$ or $w=\mu x+cs_1+ds_2$, for some $a,b,c,d \in \mathbb{C}$, $\lambda,\mu \in \mathbb{C}-\{0\}$.
- $\mathbb{C}(s_1,s_2)=\mathbb{C}(x+y,xy)$ (thanks to Peter Wu's comments).
Several ideas:
(i) First, it is convenient to concentrate on the special case where $w=y$ (or $w=x$), so $\mathbb{C}(s_1,s_2,y)=\mathbb{C}(x,y)$. We need to show that $x \in \mathbb{C}(s_1,s_2)$.
(ii) Actually, I have listed several low degree symmetric elements $x+y,xy,x^2+y^2,x^2y+xy^2,x^3+y^3$ etc. and checked for each two chosen $s_1,s_2$ from that list when, together with $y$, the three generate $\mathbb{C}(x,y)$; always one of $s_1,s_2$ was $x+y$ or $xy$.
(iii) It is clear that if $s=x+y$ or $s_1=xy$, then regardless of what $s_2$ is, we have $\mathbb{C}(s_1,s_2,y)=\mathbb{C}(x,y)$.
(iv) $\mathbb{C}(s_1,s_2) \subseteq \mathbb{C}(x+y,xy) \subseteq \mathbb{C}(x,y)$. Therefore, the degree of the field extension $\mathbb{C}(s_1,s_2) \subseteq \mathbb{C}(x,y)$ is a multiple of $2$.
(v) Probably, a proof for my above claim would involve arguments similar to the ones appearing in this answer.
(vi) It seems quite difficult for two arbitrary symmetric $s_1,s_2$ and one arbitrary non-symmetric $w$ to generate $\mathbb{C}(x,y)$, even when $w=y$! For example: $s_1=x^2+y^2,s_2=x^3+y^3,w=y$ generate $\mathbb{C}(x,y)$, but $s_1=x^2+y^2,s_2=x^4+y^4,w=y$ do not generate $\mathbb{C}(x,y)$ it generates $\mathbb{C}(x^2,y)$.
Thank you very much!