I have recently read an interesting article about the Jacobian Conjecture, in particular the reduction to the case $f(x) = x + A(x)^3$.

I was wondering about codimension one divisors on $Y = A^n$. Let $f$ be as above between $X = A^n \to Y = A^n$.

Let $T$ be a divisor on $Y$ and let $\mathscr{O}$ be the valuation ring associated to $T$ inside $K(Y)$. As I understood, when normalizing $\mathscr{O}$ inside $K(X)$ one obtains new valuation rings $\mathscr{O}_1,..,\mathscr{O}_m$, and each map to a point inside $\mathbb P^n$, the projective closure of $X$ (from the valuative criterion).

My question is this: Everytime I try to blowup closed points inside $\mathbb P^n$ for concrete examples of $A$, and study how the map $f(x)$ looks on the blowup, it seems, that the new divisors introduced by the blowups always map to the plane at infinity in $\hat{Y} = \mathbb P^n$, the projective closure of $Y$. Of course, it could be that f is not defined at the images of the $\mathscr{O}_i$ in $\hat{X} = \mathbb P^n$, but this is the reason why I blow up.

Does anybody have an example of a matrix $A$, and a divisor $T$ inside $Y$, such that one of the $\mathscr{O}_i$ is not contained in $X$, i.e. that $f: X \to Y$ is not finite over $T$? I dont claim that this will lead to new insight, but it made me very curious. Would really appreciate an example!