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 $W \in \mathbb{C}[x,y]$).


A relevant paper, for example, is [this][1].
A relevant question is [this][2].

Thank you very much! 


  [1]: https://arxiv.org/pdf/alg-geom/9305008.pdf
  [2]: https://math.stackexchange.com/questions/2803632/a-sufficient-and-necessary-condition-for-mathbbcfx-gx-mathbbcx?noredirect=1&lq=1