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A rational map $X \to Y$ is an equivalence class of maps defined on open subsets of $X$, modulo coincidence on open subsets. A birational morphism/map is a morphism which has the property that its restriction on some open subset is an isomorphism between open subsets of $X$ and $Y$. So, a rational map is not really a map, but a birational map is a real map. Of course, a birational map defines a rational map (which is invertible as a rational map; by the way, any rational map is a composition of an inverse of a birational map and a regular one).

A rational map can also be defined via its graph: it is a closed subscheme $\Gamma$ of $X \times Y$ such that for some open $U \subset X$ the projection on $U$ induces an isomorphism. There is a maximal such open $U$, the complement of it is called the set of fundamental points (or fundamental domain or indetemrinancy locus). The map is a real map on $U$, however, on the fundamental points it is a multi-valued map. An image of a fundamental point is then some set, it is obtained by composing $p_1^{-1}$ and $p_2$. It is called a total transform.

Zariski wanted to prove something about birational maps. Namely

Theorem. (original ZMT, I) Let $f: X \to Y$ be a birational map of projective varieties, and assume $X$ is normal. If x \in X is a fundamental point then its total transform is connected (and of dimension $> 0$).

Sometimes it is stated a bit more generally:

Theorem. (original ZMT, II) Let $f: X \to Y$ be a birational projective map of Noetherian integral schemes and assume that $Y$ is normal. Then for every $y \in Y$ $f^{-1}(y)$ is connected. (another verison is that you do not suppose the whole $Y$ to be normal, but just the local ring of your point $y$)

To get ZMT-I, apply ZMT-II it to the $p_1$ of the graph of $f$.

In EGA III there are two sections called  "théorème de connexion de Zariski" (EGA III, 4.3) and "le Main Theorem de Zariski (EGA III, 4.4). The key statements are

Theorem. (4.3.1, Stein factorisation) Let $f: X \to Y$ be a proper morphism, $Y$ locally Noetherian. Then $f$ factors into a compositon of a proper morphism with connected fibres and a finite morphism.

Theorem. (4.4.1, "Main Theorem") Let $Y$ be a locally Noetherian prescheme and $f:X \to Y$ be a proper morphism. Let $X′$ be the set of points in $X$ which are isolated in their fibre $f^{−1}(f(x))$. Then $X′$ is an open subset of $X$ and if $f = g \circ f′: X \to Y′ \to Y$ is the Stein factorisation of f, the restriction of $f′$ to $X′$ is an isomorphism of $X′$ onto a induced subscheme of on an open $U \subset Y′$.

These theorems together imply the original ZMT.

In EGA IV, Grothendieck proves a similar statement ("something factors into an open immersion and and a finite morphism")

Theorem. (ZMT in Grothendieck's form) Let $f: X \to Y$ be a separated quasi-finite map and $Y$ be quasi-compact. Then $f$ factors into a composition of an open immersion and a finite morphism.

It is a bit confusing why this is a "Zariski's Main Theorem" as it doesn't say anything about connectedness. However, if $f$ is projective then the set of points isolated in their fibre is quasi-finite, so by applying this theorem we get EGA III, 4.4.1. On the other hand if one supposes $f$ to be quasi-projective and quasi-finite, the theorem follows by EGA III, 4.4.1.

Anyway, it's a powerful structure theorem and it's precisely what we need for our purposes.

Proposition. Let $f: X \to Y$ be a bijective morphism of varieties over an algebraically closed field. Suppose that $Y$ is normal.  If char=0 then it is an isomorphism, if char > p then it is an isomorphism possibly composed with a power of the Frobenius.

If the extension is non-trivial and separable, take an element $h \in k(Y), a \notin k(X)$. Then the minimal polynomial $P(z)$ of $h$ will have distinct roots. The discriminant of $P(z)$ is therefore not identically zero. Then there will be an open subset of the set $U \subset Y$ where $h$ is regular and where $h$ takes more than one value. Hence $f$ is not bijective, contradiction.

Othewise, if the extension is trivial then we are done, as $k[Y]$ is integrally closed in $k(Y)=k(X)$ and $k[X]$ is an integral extension.

Since any field extension is a sequence of a separable and purely inseparable extensions, the only case left is a purely inseparable extension. In this case $f$ is just a power of the Frobenius morphism.