Let $I$ be a maximal ideal of $\mathbb{C}[x,y]$. It is well-known that $I$ is of the following form $\langle x-\alpha, y-\beta \rangle$, where $\alpha,\beta \in \mathbb{C}$.
Assume that $\langle u,v \rangle$ is a maximal ideal of $\mathbb{C}[x,y]$.
Is it true that $u=g(x), v=g(y)$, where $g$ is an automorphism of $\mathbb{C}[x,y]$?
For example, $u= x+y^3, v=y$ generate the maximal ideal $\langle x,y \rangle$ (here $\alpha=\beta=0$ and $u,v$ are images of the automorphism $g: (x,y) \mapsto (x+y^3,y)$ of $\mathbb{C}[x,y]$).
Thank you very much!
Consider $\langle u,v\rangle$ a maximal ideal of $\Bbb{C}[X,Y]$.
We have
$$\begin{align} u(x,y)&=p(x,y)(x-\alpha)+q(x,y)(y-\beta)\\ v(x,y)&=r(x,y)(x-\alpha)+r(x,y)(y-\beta) \end{align}$$
Can this transformation be an automorphism of $\Bbb{C}[X,Y]$ as a vector space (it is not an automorphism of algebra)
The question is what is the value of the determinant
$$\begin{vmatrix} p(x,y) & q(x,y)\\ r(x,y) & s(x,y) \end{vmatrix}$$
More specifically can it be $0$ if the ideal is maximal.
Let’s look at an example
Consider $p(x,y)=x=q(x,y)$ and $r(x,y)=y=s(x,y)$ we have:
$$\begin{align} u(x,y)&=x^2+xy-(\alpha+\beta)x\\ v(x,y)&=y^2+xy-(\alpha+\beta)y \end{align}$$
It seems to me that this is not an automorphism and that the ideal is the same