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Let $k$ be a field of characteristic zero.

Assume that $f: k[x_1,x_2] \to k[x_1,x_2]$ is a $k$-algebra endomorphism having an invertible Jacobian, namely, if we denote $f_1=f(x_1), f_2=f(x_2)$, then $(f_1)_x(f_2)_y-(f_2)_x(f_1)_y \in k^*$. Also assume that the leading term of $f_1$ with respect to the $x_1$-degree is $\lambda x_1^n$ (for some $\lambda \in k^*$ and $n \geq 1$) namely, $f_1= \lambda x_1^n+x_1^{n-1}g_{n-1}(x_2)+\cdots+x_1g_1(x_2)+g_0(x_2)$, where $g_{n-1}(x_2),\ldots,g_1(x_2),g_0(x_2)$ are polynomials in $x_2$ over $k$.

I wish to show that such $f$ is an automorphism. My proof is as follows; I just write $f_2= x_1^m h_m(x_2)+x_1^{m-1} h_{m-1}(x_2)+\cdots+x_1h_1(x_2)+h_0(x_2)$, where $h_m(x_2),h_{m-1}(x_2),\ldots,h_1(x_2),h_0(x_2)$ are polynomials in $x_2$ over $k$.

A direct computation of the Jacobian shows that the leading term (w.r.t. the $x_1$-degree) is $\lambda n x^{n+m-1} h_m'$ ($h_m'$ is the derivative of $h_m$ w.r.t. $x_2$). So, since $\lambda n \neq 0$, we must have $h_m' = 0$, so $h_m= \mu \in k^*$.

And then we continue with lower degrees.

(If we allow $n=0$, then $(g_0)_{x_2} (f_2)_{x_1} \in k^*$, so $(g_0)_{x_2}, (f_2)_{x_1} \in k^*$ hence $f_1= g_0=ax_2+b$ and $f_2=-(1/a)x_1 + t(y)$ (id the Jacobian is $1$), and such $f$ is a triangular automorphism).

My question(s):

(1) I guess there is a nicer/shorter way to show that such $f$ is an automorphism (probably, it relies on notions brought in this paper); it would be great if someone can bring another proof.

(2) Same question for the first Weyl algebra; again I computed the commutator etc. (Should I bring it as a separate question?).

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    Is it clear to you how to reduce it to lower degrees as you say? In general, it is not clear, since any polynomial can be assumed to be like the one you write by a change of variables and then Jacobian problem would have been solved. So, please explain carefully what you mean by continue with lower degrees. – 2017-02-13
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    By "continue with lower degrees" I meant that we consider the coefficient of $(x_1)^i$ (which is a polynomial in $x_2$), and get the $h_j$'s as polynomials in the $g_j$'s (and their derivatives). However, although $h_m = \lambda \in k$ and $h_{m-1}= a g_{n-1} + b$ for some $a,b\in k$, the $h_j$ with $j \leq m-2$ are much more complicated (unless $n=m$), so I do not have a proof that $f$ is an automorphism. I have not noticed that any endomorphism having Jacobian $\in k^*$ is of this form (I clearly see this now), so my question is equivalent to asking for an answer to the two-dimensional JC.. – 2017-02-13
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    Of course I did not mean to ask that... I have just calculated something and arrived at the above form and erroneously thought it was easy to show that such a form is an automorphism... Thank you Mohan for your comment. – 2017-02-13
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    I suggest you read Abhyankar's lectures at TIFR. These kind of arguments are clearly done there and in particular, he proves that JC condition implies the leading form is always of the form $x^iy^j$ (after linear change of variables) and if you can always show that it is of the form $x^n$, the JC would be solved. – 2017-02-13

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