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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.

 Proof. Regroup $x_1, \ldots, x_n$ so that $ K_0=k(x_1, \ldots, x_r)$ is a transcendental extension of $ k$ and $ K_{i-r}=K_0\langle x_{r+1}, \ldots, x_{i-1} \rangle$, $ x_i$ algebraic over $ K_{i-r}$ for $ i > r$. Let $ P_i(x)$-s be minimal polynomials of $ x_i$-s. They are polynomials over $ K_0$, i.e. have rational functions of $ x_1,\ldots,x_r $ as their coefficients, so there is a least common denominator of these, a polynomial $ g \in k[x_1,\ldots, x_r]$, such that $ P_i$ has coefficients in the ring $B=k \langle x_1,\ldots, x_r, g^{-1} \rangle$. Any element of the field $K$ has the form

$$ \sum b_\alpha\displaystyle\mathop{\prod}_{r+1 \leq e_i \leq d_i} x^{e_i}$$

where $\alpha$ is a multi-index, and $d_i$ are degrees of polynomials $P_i$, and $b_\alpha \in B$. From this one can see that if $I$ is a proper ideal of $ B$ (as a ring), then $ IK$ is a proper ideal of the field $ K$. Since $ I$ can only be trivial, $ B$ is a field. In other words it contains every rational function of $ x_1, \ldots, x_r, g^{-1}$, which cannot be true. ☐

Corollary. If $ A$ is finitely generated over an algebraically closed field then $ \mathrm{Max}_k A=\mathrm{Max} A$. ☐

Corollary. [Nullstellensatz] Let $ k$ be an a.c.f. and $ \mathfrak{a} \subset k[x_1,\ldots,x_n]$ be an ideal. Then $V(\mathfrak{a})$ is empty iff $ \mathfrak{a}=k[x_1,\ldots,x_n]$.

Model-theoretical proof of Nullstellensatz.

Theorem. [Nullstellensatz, weaker form. Together with Hilbert's basis theorem implies the version in the previous section] Let $ k$ be an algebraically closed field, $ I \subset k[x_1, \ldots,x_n]$ a proper ideal, generated by polynomials $ f_1, \ldots,f_m$. Then these polynomials have at least one common zero in $k^n$. Proof. First prove that

Claim. $ACF_p$ has quantifier elimination.

Suffices to construct a family of partial isomorphisms $ I$ between finitely generated substructures of two sufficiently saturated models $ ACF_p \models k_1, k_2$. $ I$ is non-empty: there is at least the partial isomorphism that identifies prime fields of $ k_1$ and $ k_2$. Construct the rest by induction. Let $ f$ be a partial isomorphism between substructures $ R_1 \subset k_1$, $ R_2 \subset k_2$ (remember, the language is just the language of rings) related by the family of partial isomorphisms. Let $ L_1$, $ L_2$ be the fraction fields of $ R_1$, $ R_2$. Pick an element $ a \in k_1-L_1$. There are two cases to consider:

• $ a$ is algebraic over $ L_1$ ; then pick the minimal polynomial of $ a$ and extend $ f$ by putting $ f(a)=b$ where $ b$ is a solution of the minimal polynomial of $ a$ in $ k_2$;
• $ a$ is transcendental over $ L_1$; one can choose any transcendental (over $ L_2$) element $ b$ of $ k_2$ and extend $ f$: $ f(a)=b$. The question is if there is any such element. Since any polynomial only has finitely many roots, the set of formulae $ \{ P(\bar{x}) \neq 0 \mid P \in L_2[\bar{x}]\}$ is finitely satisfiable (and is a type over a finite set, since $ L_2$ is finitely generated). By saturatedness (in fact $ \aleph_0$-saturatedness would saffice), it has a solution in $ L_2$.

Now take the smallest prime ideal $ J$ that contains $ I$. It is not the whole of $ k[\bar{x}]$ unless $ I$ is. Let $ L$ be the field of fractions of the quotient $ k[x_1, \ldots, x_n]/J$, then $ k$ canonically identifies with a substructure of $ K$. Clearly, $ f_i(\bar{x}) = 0$ in $ K$, and all the more in the algebraic closure of $ K$, $ \bar{K}$. By quantifier elimination, $ k$ is an elementary substructure of $ \bar{K}$ and so has a solution to $ f_1,\ldots,f_n$. ☐