Invertibility of a polynomial map equivalent to a condition on the ideal generated by the coordinates? Let $k$ be an algebraically closed field.
Let $$F=(F_1,\ldots,F_n) \colon k^n \to k^n$$
be a polynomial map. 
I'm trying to understand the relation between the conditions:


*

*$F$ is invertible.

*$\langle F_1,\ldots, F_n \rangle = I(F^{-1}(0))$ is a maximal ideal.


If $F$ is invertible, then $F^{-1}(0)\in k^n$ is a singleton, and the corresponding ideal is maximal. It equals the radical of $\langle F_1,\ldots, F_n \rangle$, but does it equal $\langle F_1,\ldots, F_n \rangle$ itself ?
How about the converse ?
 A: A morphism of affine varieties is invertible if and only if it the associated map of affine coordinate rings is an isomorphism. In your case, this says that $F: k^n \rightarrow k^n$ is invertible if and only if the map $F^*: k[x_1,\dots,x_n] \rightarrow k[x_1,\dots,x_n]$ given by $g(x_1,\dots,x_n) \mapsto g(F_1,\dots,F_n)$ is a ring isomorphism.
In particular, if $F$ is invertible, the image of the ideal $\langle x_1,\dots,x_n\rangle$ under $F^*$ is maximal. This image is precisely the ideal $\langle F_1,\dots,F_n\rangle$.
I claim a partial converse: if $\langle F_1,\dots,F_n\rangle$ is maximal, then $F^*$ is injective. To see this, note that injectivity is equivalent to $F_1,\dots,F_n$ being algebraically independent over $k$. By Noether's normalization, algebraic dependence of $F_1,\dots,F_n$ would imply that $k[x_1,\dots,x_n]$ is integral over $k[y_1,\dots,y_r]$ for some $y_i$ and $r<n$. This in turn would imply that the transcendence degree of $k(x_1,\dots,x_n)$ over $k$ is less than $n$, a contradiction. 
Note: In my original answer, I carelessly asserted $$\langle x_1,\dots,x_n\rangle = \langle F_1,\dots,F_n\rangle \implies k[x_1,\dots,x_n]=k[F_1,\dots,F_n].\,\,\, (*)$$ This would imply $\langle F_1,\dots,F_n\rangle$ maximal iff $F^*$ is an isomorphism. After seeing this related post, I realized $(*)$ is false: the subalgebra generated by a set of elements need not contain their ideal. As @GeorgesElencwajg points out in his answer to the referenced question, a counter example is given by $\langle y, x+xy \rangle = \langle x,y\rangle$, since $k[x,y] \neq k[y,x+xy]$.
