Inverse of a polynomial map Ax-Grothendieck Theorem states that if $\mathbf K$ is an algebraically closed field, then any injective polynomial map $P:\mathbf K^n\longrightarrow \mathbf K^n$ is bijective.
Question 1. What does the inverse map of $P$ look like ? What kind of map is that ?
$P^{-1}$ need not be polynomial, as the example $x^p$ in $\mathbf F_p^{alg}$ shows.
Question 2. Are there conditions under which $P^{-1}$ is polynomial ?
 A: In light of abx's comment, I came up with the following argument. I'm sure some version of this must be in the literature somewhere.
Recall that a dominant morphism $Y \to X$ of varieties is separable if $K(X) \to K(Y)$ is a separable field extension. In characteristic $0$, this is automatic.

Theorem. Let $k$ be an algebraically closed field, and let $f \colon Y \to X$ be a morphism of $k$-varieties, with $X$ normal. If $f$ is bijective (in particular, dominant) and separable, then $f$ is an isomorphism.

Proof. By a suitable version of Zariski's main theorem (see e.g. Tag 05K0), there exists an open immersion $Y \subseteq Z$ and a finite morphism $g \colon Z \to X$ extending $f$. By restricting to the underlying reduced scheme of the unique irreducible component of $Z$ containing $Y$, we may assume $Z$ is integral. In particular $K(Z) = K(Y)$, and $g$ is separable since $f$ is.
Since $g$ is separable, it is generically étale, hence there exists an open $U \subseteq X$ such that
$$g \colon g^{-1}(U) \to U$$
is finite étale. In particular, $g$ is finite flat over $U$ of some rank $r \geq 1$. If $x \in U$ is a closed point, then $f^{-1}(x) \to x$ is finite étale of rank $r$, thus a disjoint union of $r$ copies of $\operatorname{Spec} k$ (since $k$ is algebraically closed). Since $f$ is bijective, we have $r = 1$, so $f$ is birational, i.e.
$$K(X) \stackrel \sim \to K(Z).$$
Since $X$ is normal and $g$ finite, this forces $g$ to be an isomorphism. Since $f$ is bijective, we conclude that $Y = Z$ and $f = g$. $\square$
Remark. I do not know how to get rid of the algebraically closed condition without weakening the hypotheses to $f$ being universally bijective (in a restricted sense: we only need it for base change along field extensions) and $X$ geometrically normal (when $k$ is perfect, this is equivalent to normal).
A: $P$ has a polynomial inverse implies that the Jacobian of $P$ is a constant function. There is a conjecture known as the Jacobian conjecture which says that if the characteristic of $K$ is zero, $P$ has a polynomial inverse if and only if its Jacobian is a non zero constant.
https://en.wikipedia.org/wiki/Jacobian_conjecture
