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 $O$ be the valuation ring associated to $T$ inside $K(Y)$. As I understood, when normalizing $O$ inside $K(X)$ one obtains new valuation rings $O_1,..,O_m$, 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 $\bar{(Y)} = \mathbb P^n$, the projective closure of $Y$. Of course, it could be that f is not defined at the images of the $O_i$ in $\bar{(X)} = \mathbb P^n$, 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 $O_i$ 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!

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!

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 -> Y = A^n.

Let T be a divisor on Y and let O be the valuation ring associated to T inside K(Y). As I understood, when normalizing O inside K(X) one obtains new valuation rings O_1,..,O_m, and each map to a point inside P^n, the projective closure of X (from the valuative criterion).

My question is this: Everytime I try to blowup closed points inside 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) = P^n, the projective closure of Y. Of course, it could be that f is not defined at the images of the O_i in hat(X) = 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 O_i is not contained in X, i.e. that f: X -> 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!

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 $O$ be the valuation ring associated to $T$ inside $K(Y)$. As I understood, when normalizing $O$ inside $K(X)$ one obtains new valuation rings $O_1,..,O_m$, 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 $\bar{(Y)} = \mathbb P^n$, the projective closure of $Y$. Of course, it could be that f is not defined at the images of the $O_i$ in $\bar{(X)} = \mathbb P^n$, 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 $O_i$ 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!