Let $p,q \in \mathbb{C}[x,y]$ with $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in \mathbb{C}^{\times}$. Assume that $\deg(p)=\deg(q)$ (the total degree, also called the $(1,1)$-degree).
Claim: In this special case (equal degrees) it follows that $f: (x,y) \mapsto (p,q)$ is an automorphism of $\mathbb{C}[x,y]$.
Please, I would like to make sure that I am not missing something, and the above claim (which I have not seen in articles) is indeed true.
If $f: (x,y) \mapsto (p,q)$ is not an automorphism of $\mathbb{C}[x,y]$, then (by a result that I think I have seen in van den Essen's book, perhaps due to a result of Nagata and Lang) there exists an automorphism $g$ of $\mathbb{C}[x,y]$ such that:
- $\deg(p)=\deg(g(p))$.
- $\deg(q)=\deg(g(q))$.
- $\deg(g(p))=(a+b)n > (a+b)m=\deg(g(q))$
(since the $(1,1)$-leading term of $g(p)$ is $\lambda x^{an}y^{bn}$ and the $(1,1)$-leading term of $g(q)$ is $\mu x^{am}y^{bm}$, $\lambda,\mu \in k^{\times}$).
Then $\deg(p)>\deg(q)$, contrary to our assumption.
Remark: Now I see that there is a problem with my claim, since every endomorphism we can bring to a form with equal degrees (just multiply with an appropriate automorphism such as $(x,y) \mapsto (x+y,x-y)$), and then it would follow that the JC is true.. this seems too easy to be a valid proof for the two-dimensional JC. So the quoted result (concerning the existence of such $g$ etc.) should be wrong? (Probably I misunderstood the quoted result).
Any hints and comments are welcome!