# Concerning certain Keller maps of $k[x,y]$

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!

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