Automorphisms of $\mathbb{A}^{n}$

Question

Let $\varphi: \mathbf{A}^{n} \rightarrow \mathbf{A}^{n}$ be a morphism of $\mathbf{A}^{n}$ to $\mathbf{A}^{n}$ given by $n$ polynomials $f_{1}, \ldots, f_{n}$ of $n$ variables $x_{1}, \ldots, x_{n} .$ Let $J=\operatorname{det}\left|\partial f_{i} / \partial x_{j}\right|$ be the Jacobian polynomial of $\varphi .$

show that If $\varphi$ is an isomorphism (in which case we call $\varphi$ an automorphism of $\mathbf{A}^{n}$ ) show that $J$ is a nonzero constant polynomial.

solution :

If $\varphi \in \operatorname{Aut}\left(\mathbb{A}^{n}\right),$ then each $f_{i} \notin k,$ since then $\varphi$ is not surjective. Therefore each $f_{i}$ is a linear non-constant polynomial, so $J \in k^{\times}$

why $f_{i} \notin k$ ? why each $f_{i}$ is a linear non-constant polynomial ? also why $J \in k^{\times}$ ?

Answer 1

• $J\in k[x_1,\ldots,x_n]$

• $J$ doesn't vanish on $\overline{k}^n$

• Thus $J$ is constant, it is in $k^\times$

For $n=2$ try with $\varphi(x_1,x_2)=(x_1+x_2^2,x_2)$ to see that $Aut(\Bbb{A}^n)$ is much more complicated than $Aut(\Bbb{P}^n)$ where we get "it must be linear".

The Jacobian conjecture is the converse: if $J\in k^\times$ then does $\varphi$ have to be an automorphism?

Answer 2

Hint Try using the inverse polynomial map and the chain rule. What are the invertible elements of a polynomial ring?