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

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 $$l_{0,1}(A)=-7y^6$$.

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.

• @ThomasAndrews, thank you for the comment. I will edit my question. Commented Jul 17, 2019 at 18:49
• Oh, the scalar jacobian is a condition. I get it now, sorry. Commented Jul 17, 2019 at 18:50
• Still not clear what $l_{1,-1}$ and $\deg_{1,-1}$ mean. Commented Jul 17, 2019 at 18:51
• ok. (I added an exampe). I will add the definition of $l_{1,-1}$. Commented Jul 17, 2019 at 18:52
• If I'm reading this correctly, you can assume $\lambda=\mu=1$ to make it potentially less messy. Commented Jul 17, 2019 at 18:58

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.

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

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:

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

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.

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.

• Thank you very much! How to obtain $p=x$, $q=x+y$? Commented Jul 17, 2019 at 20:17
• Oh, I see. We take $n=0$, $t=1$, $\beta=0$, $J=-1$, $\alpha=\gamma=1$, and get $p=x+y$ and $q=y$. Commented Jul 17, 2019 at 20:20
• I don't think I get all of them - it is actually the case that $b_y-a_y$ must be of the form $\delta x^{1-n}$ for some $\delta \in k.$ I only used used $\delta=0.$ @user237522 Commented Jul 17, 2019 at 20:26
• The case when $\delta\neq 0$ is really hairy. Commented Jul 17, 2019 at 20:55
• Thank you. Any further comments/elaboration on the $\delta \neq 0$ case is welcome. Commented Jul 18, 2019 at 11:59