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!

  • $\begingroup$ 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 $1/y$. $\endgroup$ – LSpice May 18 at 1:33
  • $\begingroup$ @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 $s=\frac{a}{b}$. $\endgroup$ – user237522 May 18 at 4:09
  • $\begingroup$ 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). $\endgroup$ – user237522 May 18 at 4:15
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    $\begingroup$ 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 $y$. $\endgroup$ – user237522 May 18 at 4:25

The answer is no.

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 $\deg(p)\ge 2$.

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  • $\begingroup$ Thank you very much! $\endgroup$ – user237522 Jul 9 at 12:17

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