Injectivity of Keller maps

Question

Let $M: \mathbb{C}[x,y] \to \mathbb{C}[x,y]$, $(x,y) \mapsto (p,q)$, with $p,q \in \mathbb{C}[x,y]$ satisfying $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in \mathbb{C}-\{0\}$. Such a polynomial map is called a Keller map, and the two-dimensional Jacobian Conjecture says that such a map is injective and surjective.

I know that there are many papers trying to prove injectivity or surjectivity of Keller maps.

My question is quite basic, and I think it is dealt with in one paper or another or perhaps in algebraic geometry books, but I am not able to find the relevant references now.

Further assume that $p(0,0)=q(0,0)=0$.

Question: If $M$ is not injective, then is it true that there exist $c,d \in \mathbb{C}-\{0\}$ such that $p(c,0)=q(c,0)=0$ and $p(0,d)=q(0,d)=0$?

I ask, in other words: Given a map $\tilde{M}: \mathbb{C}^2 \to \mathbb{C}^2$, defined by $(a,b) \mapsto (p(a,b),q(a,b))$ (where $p,q \in \mathbb{C}[x,y]$ have invertible Jacobian), if we assume that $\tilde{M}$ is not injective and $(0,0) \mapsto (0,0)$, is it true that there exist $c,d \in \mathbb{C}-\{0\}$, with $(c,0) \mapsto (0,0)$ and $(0,d) \mapsto (0,0)$.

(What I remember vaguely is something like: 'If $M$ is not injective, then it is not injective at zero'; I am not sure about my specific question. I guess it was meant that instead of $M$ we can change variables= compose it with an automorphism, and then get what I asked?).

What I have tried: Write: $p=p_ny^n+\cdots+p_1y+p_0$, where $p_i \in k[x]$ and $q=q_my^m+\cdots+q_1y+q_0$, where $q_j \in k[x]$. The assumption $p(0,0)=q(0,0)=0$ implies that $p_0(0)=0$ and $q_0(0)=0$. We have: $p(x,0)=p_0$ and $q(x,0)=q_0$. I am asking if $p_0$ and $q_0$ have a common root $c \in \mathbb{C}-\{0\}$ other than $0$ ($0$ is a common root of $p_0$ and $q_0$). We assumed that $p_0=xf$ and $q_0=xg$, for some $f,g \in k[x]$. So I ask if $f$ and $g$ have a common root $c \in \mathbb{C}-\{0\}$. Observation: $f$ and $g$ cannot have $0$ as a common root, since in this case $p_0=x^2\tilde{f}$ and $q_0=x^2\tilde{g}$, for some $\tilde{f},\tilde{g} \in k[x]$. But then $\operatorname{Jac}(p,q)|(0,0)=0$ ($p_x(0,0)=0$ and $q_x(0,0)=0$). Therefore, $f$ and $g$ are coprime or have a common root other than $0$. If $f$ and $g$ are coprime, then $\gcd(p(x,0),q(x,0))=\gcd(p_0,q_0)=x$. (Similarly, $p(0,y)$ and $q(0,y)$ have $0$ as a common root, and if they do not have other than $0$ common roots, then $\gcd(p(0,y),q(0,y))=y$). Then $p=p_ny^n+\cdots+p_1y+xf$, $q=q_my^m+\cdots+q_1y+xg$, with $\gcd(f,g)=1$. (There exist $u,v \in k[x]$ such that $uf+vg=1$. Then, $up+vq=Wy+x$, for some $W \in \mathbb{C}[x,y]$).

Relevant papers, for example, are Injectivity on one line and One-one polynomial maps (Lemma 1). A relevant question is this.

Now asked also in MSE.

Thank you very much!

Answer 1

This can be achieved after appropriate changes of coordinates of the source and target. More precisely, there are automorphisms $A, B$ of $\mathbb{C}[x,y]$ such that $A \circ M \circ B$ has the property you want.

This follows e.g. from Orevkov's result that if $\tilde M: \mathbb{C}^2 \to \mathbb{C}^2$ is a counterexample to the Jacobian conjecture, then the topological degree $d$ of $\tilde M$ is $\geq 4$. For generic $c \in \mathbb{C}^2$, $\tilde M^{-1}(c)$ has $d$ points.

Claim: You can choose $c$ such that not all points of $\tilde M^{-1}(c)$ are collinear.

(You are done if the claim holds, since then by changing coordinates you can map $c$ to the origin and $3$ non collinear points on $\tilde M^{-1}(c)$ respectively to the origin, a point on the $x$-axis and a point on the $y$-axis.)

There must be an elementary way to prove the claim for more general maps, but now I can only think of the following argument which applies only to non-injective Keller maps $\tilde M$: due to the "injectivity on one line" result you mentioned, the restriction of $\tilde M$ to each "vertical line" $y = b$ is non-injective. Moreover, for almost all $b \in \mathbb{C}$, the degree of $\tilde M|_{y = b}$ is smaller than the global degree $d$ (since otherwise $\tilde M|_{x = a}$ will be injective for all $a$). Therefore we can choose points $(a_1, b), (a_2, b), (a', b') \in \mathbb{C}^2$ which map to the same point under $\tilde M$ and such that $a_1 \neq a_2$ and $b \neq b'$, as required.