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Let $k$ be a field of characteristic zero.

Let $(x,y) \mapsto (p,q) \in k[x,y]$ be a Keller map, namely, a $k$-algebra endomorphism of $k[x,y]$ with $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in k-\{0\}$.

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 $j$, $0 \leq j \leq n$. $A_n$ 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 $l_{0,1}(A)=-7y^6$.

Question 1: Is it possible to find all Keller maps satisfying the following two conditions: (i) Each of $\{l_{1,-1}(p),l_{1,-1}(q)\}$ is a monomial. (ii) $l_{1,-1}(p)+l_{1,-1}(q)=0$.

Of course, the identity map $(x,y) \mapsto (x,y)$ is a Keller map satisfying (i) and (ii). What about other examples? It seems that there exist no other examples; am I missing something?


More generally,

Let $(x,y) \mapsto (p,q) \in k[x,x^{-1},y]$ be a generalized Keller map, namely, a $k$-algebra homomorphism from $k[x,y]$ to $k[x,x^{-1},y]$ with $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in k-\{0\}$.

Question 2: Is it possible to find all generalized Keller maps satisfying the following two conditions: (i) Each of $\{l_{1,-1}(p),l_{1,-1}(q)\}$ is a monomial. (ii) $l_{1,-1}(p)+l_{1,-1}(q)=0$.

An example: $p=x^2y+3+x^{-1}+x^{-4}+2x^{-15}$, $q=x^{-1}$.

Question 3: (1) Are there other types of examples except $p=x^{1+m}y+T$, $q=x^{-m}$, where $T \in k[x,x^{-1}]$ has an 'appropriate' $(1,-1)$-degree? (2) If we assume that the $(1,-1)$-leading terms are $l_{1,-1}(p)= x^{1+m}y$, $l_{1,-1}(q)=x^{-m}$, is it true that necessarily: $p=x^{1+m}y+T$, $q=x^{-m}$, with appropriate $T \in k[x,x^{-1}]$?

Same questions with $(x,y) \mapsto (p,q) \in k[x^{1/r},x^{-1/r},y]$, $r \in \mathbb{Z}$.


I have asked the above question here.

Edit: If I am not wrong, I have a proof for the following similar claim:

Let $U,V$ be two $(1,-1)$-homogeneous elements of $k[x^{1/r},x^{-1/r},y]$ (not necessarily monomials), having a non-zero scalar Jacobian, namely, $\operatorname{Jac}(U,V)\in k-\{0\}$.

Then $\{U,V\}$ is one of the following sets:

(1) $E_1:=\{\lambda x,\mu(y+x^{-1})\}$.

(2) $E_2:=\{\lambda x^{1+l/r}y, \mu x^{-l/r}\}_{l \neq 0}$.

A sketch of proof:

Write $U=u_nx^{a}y^{b}+u_{n-1}x^{a-1}y^{b-1}+\cdots+u_{s}x^{a-(n-s)}y^{b-(n-s)}$ and $V=v_mx^{a}y^{b}+v_{m-1}x^{a-1}y^{b-1}+\cdots+v_{t}x^{a-(m-t)}y^{b-(m-t)}$, where $a,c \in \mathbb{Z}/r$, $b,d \in \mathbb{N}$.

We have assumed that $\operatorname{Jac}(U,V)\in k-\{0\}$, therefore $(a-b)+(c-d)=0$ (see Remark 1.12 of this paper).

Expand $\operatorname{Jac}(U,V)$.

By considerations of $(1,1)$-degrees (yes, $(1,1)$-degrees, not $(1,-1)$-degrees, which are all equal, since this Jacobian is $(1,-1)$-homogeneous of $(1,-1)$-degree $(a-b)+(c-d)$), we obtain that $(b+d)(b-a)=0$ or several low degrees cases.

It is possible to show that each case (= "$(b+d)(b-a)=0$ or several low degrees cases") implies that $\{U,V\}$ is $E_1$ or $E_2$.

Thank you very much!

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  • $\begingroup$ It would help if you recalled the definition of $(a,b)$-leading term. $\endgroup$ – Abdelmalek Abdesselam Jun 18 at 15:49
  • $\begingroup$ Thanks for the clarification. $\endgroup$ – Abdelmalek Abdesselam Jun 18 at 21:32
  • $\begingroup$ Thank you for your comment! (I was just about to inform you that I have added the definition). $\endgroup$ – user237522 Jun 18 at 21:32

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