# Certain endomorphisms of $\mathbb{C}(x,y)$

Let $$f: (x,y) \mapsto (p,q)$$ be a $$\mathbb{C}$$-algebra endomorphism of $$\mathbb{C}(x,y)$$ satisfying the following two conditions:

(i) $$\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in \mathbb{C}-\{0\}$$.

(Generally, $$\operatorname{Jac}(p,q) \in \mathbb{C}(x,y)$$).

(ii) One of $$\{p,q\}$$ can be written as $$\frac{u}{v}$$, where $$u,v \in \mathbb{C}[x,y]$$, $$\gcd(u,v)=1$$ and (exactly) one of $$\{u,v\}$$ is a multiple of $$y$$.

(Edit: The original condition (ii) was slightly different and unclear).

Question: Is such $$f$$ necessarily an automorphism of $$\mathbb{C}(x,y)$$?

Examples: $$f: (x,y) \mapsto (xy^2,\frac{1}{y})$$. We have, $$\operatorname{Jac}(xy^2,\frac{1}{y})=-1$$. It is clear that $$f$$ is an automorphism of $$\mathbb{C}(x,y)$$ (obviously, $$x$$ and $$y$$ are in the image of $$f$$).

$$g: (x,y) \mapsto (x^2,y^2)$$ is not an automorphism of $$\mathbb{C}(x,y)$$, but this does not contradict a positive answer to my question, since $$g$$ satisfies condition (ii) but does not satisfy condition (i).

Motivation: If we replace $$\mathbb{C}(x,y)$$ by $$\mathbb{C}[x,y]$$, then by a known result concerning the Newton polygon we obtain that such $$f$$ is an automorphism of $$\mathbb{C}[x,y]$$.

The known result can be found, for example, in Essen's book Proposition 10.2.6, in Cheng-Wang's paper Lemma 1.14, Nowicki-Nakai's paper Proposition 2.1 and Nagata's paper.

Remark: I suspect that the answer to my above question is yes, but I am not sure if the proof for $$\mathbb{C}[x,y]$$ can be adjusted to $$\mathbb{C}(x,y)$$.

Thank you very much!

• In (ii), what kind of multiple? I read $\mathbb C(x, y)$ as the fraction field of $\mathbb C[x, y]$, in which case every element is a multiple of $y$ and of $1/y$. – LSpice May 18 at 1:33
• @LSpice, interesting comment, thank you! What I had in mind is as follows: Given the involution $\beta: (x,y) \mapsto (x,-y)$ (on $\mathbb{C}[x,y]$ extended to $\mathbb{C}(x,y)$), according to math.stackexchange.com/questions/3569468/… if $s \in \mathbb{C}(x,y)$ is symmetric w.r.t. $\beta$ then we can find symmetric $a,b \in \mathbb{C}[x,y]$ such that $s=\frac{a}{b}$. – user237522 May 18 at 4:09
• Similarly, if $k \in \mathbb{C}(x,y)$ is skew-symmetric w.r.t. $\beta$ then we can find symmetric $c \in \mathbb{C}[x,y]$ and skew-symmetric $d \in \mathbb{C}[x,y]$ such that $k=\frac{c}{d}$ or $k=\frac{d}{c}$. (for example $k=\frac{1}{y}=\frac{y}{y^2}$). Now, I am interested in a $\mathbb{C}$-algebra endomorphism $f: (x,y) \mapsto (p,q)$ of $\mathbb{C}(x,y)$ such that (i) is satisfied, $p$ is symmetric and $q$ is skew-symmetric. From $q$ skew-symmetric I imposed condition (ii). – user237522 May 18 at 4:15
• Probably a better version of (ii) is as follows: One of $\{p,q\}$ can be written as $\frac{u}{v}$, where: $u,v \in \mathbb{C}[x,y]$, $\gcd(u,v)=1$ and (exactly) one of $\{u,v\}$ is a multiple of $y$. – user237522 May 18 at 4:25

Take $$p=\frac{x^2}{2}$$, take $$q=\frac{y}{x}$$. The Jacobian matrix is $$\begin{pmatrix} x& -\frac{y}{x^2} \\ 0 & \frac{1}{x}\end{pmatrix}$$ whose determinant is equal to $$1$$. However, $$f$$ is definitely not an automorphism of $$\mathbb{C}(x,y)$$.
More generally, take any polynomial $$p\in\mathbb{C}[x]$$ and choose $$q=\frac{y}{p_x}$$. This gives you a counterexample as soon as $$\deg(p)\ge 2$$.