4
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
5
$\begingroup$

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 $\deg(p)\ge 2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you very much! $\endgroup$ – user237522 Jul 9 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.