Form of $(x,y) \mapsto (\lambda x^n+\cdots,\mu x^n+\cdots) \in k[x^{1/m},x^{-1/m},y]$

Question

Let $k$ be a field of characteristic zero, $m \in \mathbb{Z}$, and $k[x^{1/m},x^{-1/m},y]$ the polynomial ring generated by three commuting variables $x^{1/m},x^{-1/m},y$ subject to $(x^{-1/m})^m x=1$.

Let $A \in k[x,y]$. Recall that given $(a,b) \in \mathbb{Z}^2$ (sometimes it is required that $\gcd(a,b)=1$), we can write $A=A_n+A_{n-1}+\cdots+A_1+A_0$, where $A_n \neq 0$, and $A_j$ is $(a,b)$-homogeneous of $(a,b)$-degree $j$, $\deg_{a,b}(A_j)=j$, $0 \leq j \leq n$. $A_n$ is called the $(a,b)$-leading term of $A$ and is denoted by $l_{a,b}(A)$.

For example: If $A=x^2y^2+8x^3y^3-7y^6$, then $l_{1,1}(A)=8x^3y^3-7y^6$, $l_{1,-1}(A)=x^2y^2+8x^3y^3$, $l_{1,0}(A)=8x^3y^3$ and $l_{0,1}(A)=-7y^6$.

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Let $f: (x,y) \mapsto (p,q)$ be a $k$-algebra homomorphism from $k[x,y]$ to $k[x^{1/m},x^{-1/m},y]$ having a non-zero scalar Jacobian, namely, $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in k-\{0\}$.

Assume that $l_{1,-1}(p)= \lambda x^n$ and $l_{1,-1}(q)= \mu x^n$, for some $n \in \mathbb{N}-\{0\}$, $\lambda, \mu \in k-\{0\}$. Then we can write $p=\lambda x^n+a$, $q=\mu x^n+b$, with $\deg_{1,-1}(a)< n$ and $\deg_{1,-1}(b) < n$.

For example: $p=x$ and $q=x+y$.

Is it true that such $f$ must be of the following form:

$p=\lambda x^n$ and $q=\mu x^n+H+\nu x^{1-n}y+G$, where $H,G \in k[x^{1/m},x^{-1/m}]$, $\nu \in k-\{0\}$.

(For convenience, we may assume that $\lambda=\mu=1$).

Please see this relevant question.

Any hints and comments are welcome! Thank you.

Answer 1

Edit: This answer is wrong - when I conclude that $a_y-b_y$ must be zero, we actually have that $a_y-b_y=\delta x^{1-n}$ for some $\delta\in k.$ So my conclusion only gets some of the pairs $p,q$ of this form. Working on a fix.

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You can restrict the case where $\lambda=\mu=1.$

If $p=x^n+a, q=x^n+b$ then $$ \begin{align}p_xq_y-p_yq_x&=(x^{n-1}+a_x)b_y-b_y(nx^{n-1}+b_x)\\ &=nx^{n-1}(b_y-a_y) + (a_xb_y-b_xa_y.)\end{align}$$

To get a scalar, for $n>1$ you need $b_y-a_y=0$ (sic: See note at top.) So $a=b+H$ for some $H\in k[x^{1/m},x^{-1/m}].$ Then you need $a_xb_y-b_xa_y=b_y(a_x-b_x)=b_yH$ to be a non-zero scalar. So you need $b_y\in k\left[x^{1/m},x^{-1/m}\right]$ with $b_yH$ a non-zero scalar.

\So you need $H,G\in k\left[x^{1/m},x^{-1/m}\right]$ such that $HG$ is a non-zero scalar and then $b_y=G$ so $b=Gy+J$ for some $J\in k\left[x^{1/m},x^{-1/m}\right].$

So, given $G,H,J\in k\left[x^{1/m},x^{-1/m}\right]$ such that $GH_x$ is a non-zero scalar, then $p(x)=x^n+Gy+J+H,$ and $q(x)=x^n+Gy+J.$

Then $$p_xq_y-p_yq_x=(nx^{n-1}+G_xy+J_x)G-(nx^{n-1}+G_xy+J_x+H_x)G=GH_x$$ is a non-zero scalar.

This works when $H=\alpha x^{t}+\beta$ and $G=\gamma x^{1-t}$ where $\alpha,\gamma\in k\setminus \{0\}$ and $\beta\in k,$ where $t\in\frac{1}{m}\mathbb Z\setminus 0,$ and $t<n.$

So we get that we have $$\begin{align}p&=x^n+\gamma x^{1-t}y +\alpha x^t+\beta+J,\\ q&=x^n+\gamma x^{1-t}y +J \end{align}$$

where $\alpha,\gamma$ are non-zero scalars, $\beta$ is any scalar, $t\in\frac{1}{m}\mathbb Z\setminus \{0\}$ and any $J\in k\left[x^{1/m},x^{-1/m}\right].$

Then the Jacobian is:

$$GH_x=t\alpha\gamma.$$

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This assumes that the only invertible elements in $k\left[x^{1/m},x^{-1/m}\right]$ are of the form $a(x^{1/m})^n$ where $a\in k\setminus \{0\}$ and $n$ is any integer. This is relatively easy to prove.

You might have some conditions on the size of $t$ and the degree of $J$ to keep this kosher with your question.

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The general case when $(\lambda,\mu)\neq (1,1)$ can be gotten just by putting $\lambda$ and $\nu$ as the coefficient of $x^n$ in the two polynomials.