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](https://arxiv.org/pdf/alg-geom/9305008.pdf) and [One-one polynomial maps](https://www.jstor.org/stable/2034426) (Lemma 1).
A relevant question is [this](https://math.stackexchange.com/questions/2803632/a-sufficient-and-necessary-condition-for-mathbbcfx-gx-mathbbcx?noredirect=1&lq=1).


Now asked also in [MSE](https://math.stackexchange.com/questions/4215275/injectivity-of-keller-maps).


Thank you very much!



Injectivity of Keller maps
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](https://www.math.univ-toulouse.fr/~orevkov/jc86.pdf) 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.


