Let’s first think about why complex algebraic geometry is nice. If you do topology, you have to deal with bad continuous function, and if you do algebraic geometry over $\mathbb{Q}$, you get complications very quickly, e.g., the image of $x \mapsto x^2$ is not easily understandable over $\mathbb{Q}$. On the other hand, over $\mathbb{C}$, Chevalley’s theorem tells us that the image of a map between algebraic varieties is possibly neither open nor closed, but at least a finite union of locally closed sets.
Structures and definable sets#
Definition 1. Let $M$ be a set, and let $M^\infty = \coprod_n \mathcal{P}(M^n)$. A structure on $M$ is a subset of $M^\infty$ that is closed under
- boolean operations,
- products,
- projections.
Given any subset $S \subseteq M_\infty$, there is then going to be a structure generated by $S$.
Example 2. If we start with $M = \mathbb{C}$ and consider the structure generated by $\lbrace z \rbrace$ in $\mathbb{C}$ for all $z \in \mathbb{C}$ and the graphs $\lbrace (x, y, x+y) \rbrace$ and $\lbrace (x, y, xy) \rbrace$ in $\mathbb{C}^3$, then the definable sets are precisely the constructible sets.
For $\mathbb{Q}$, the structure is disastrous, but over $\mathbb{R}$, this is reasonably nice. In particular, the relation $x \le y$ can be defined algebraically because it is equivalent to $\exists z (y = x + z^2)$.
Theorem 3 (Tarski–Seidenberg). All definable subsets of $\mathbb{R}^n$ are semi-algebraic, i.e., is in the boolean algebra generated by $p(x) = 0$ and $q(x) \gt 0$.
O-minimality#
Definition 4. Let $S$ be a structure on $\mathbb{R}$ containing all semi-algebraic sets. We say that $S$ is o-minimal when all definable subset of $\mathbb{R}$ are finite unions of points and intervals.
As seen above, the semi-algebraic sets form a o-minimal structure.
Lemma 5. Let $S$ be an o-minimal structure, and let $f \colon I = [a, b] \to \mathbb{R}$ be a definable function, i.e., its graph is definable. Then there exists a nontrivial subinterval $I^\prime \subseteq I$ where $f$ is either constant or injective.
Proof.
If a fiber $f^{-1}(r)$ is infinite, then we are done. So suppose that all fibers are infinite. Then $f(I)$ is infinite, so it contains an interval. Now we can define $g \colon J \to I$ by setting $g(j)$ to be the least $i$ such that $f(i) = j$. This is also definable, by using first order logic. Then $g(J)$ contains an interval, and then on this interval we have injectivity.
Lemma 6. With notation as above, $f$ has finitely many discontinuities.
Proof.
The formula that says “$f$ is continuous at $x$” is definable. So this is a finite union of points and intervals. Now we can show that $f$ is locally monotone except for a finite set of points, and then $f$ is continuous almost everywhere in the sense of measure theory. Now we have that the set of discontinuities of $f$ is finite.
Lemma 7 (uniform finiteness). Let $X \subseteq \mathbb{R}^2$ be a definable set and assume that all fibers for $\pi \colon \mathbb{R}^2 \to \mathbb{R}$ are finite. Then they are uniformly finite.
Proof.
We say that a point $(a, b)$ is regular when there is a box on which the projection is homeomorphism. This is a definable property, so let $B \subseteq \mathbb{R}$ be the image of the all irregular points. Now away from $B$, the function $\lvert \pi^{-1}(x) \rvert$ is locally constant. So it is enough to show that $B$ is finite. It is enough to rule out the case when $B$ contains an interval. Define a function $$ \beta \colon B \to \mathbb{R} $$ similarly so that $\beta(a)$ is the least $b$ such that $(a, b)$ is irregular. This is again definable, and it is continuous and monotone on some subinterval.
Let us assume that these $(a, \beta(a))$ are actually in $X$, and these are neither the smallest nor largest elements in the fibers $\pi^{-1}(a)$. (The other cases can be dealt with similarly, where we definably divided cases so that we have an interval loci of these cases.) Let $(a, \beta_-(a))$ be the next lower elements in the fibers, and let $(a, \beta_+(a))$ be the next higher elements in the fibers. Then these are also continuous, and now the section $(a, \beta(a))$ is sandwiched between two sections, so that this is actually regular.
Remark 8. The structure $\mathbb{R}_ {\mathrm{an},\mathrm{exp}}$ generated using polynomials, the real exponential $\exp \colon \mathbb{R} \to \mathbb{R}$, and real analytic functions on compact domains, is o-minimal.
The definable Chow theorem#
Now we can state the definable Chow theorem.
Theorem 9. If $X \subseteq \mathbb{C}^n$ is definable in an o-minimal structure and analytic, then $X$ is algebraic.
Lemma 10. Let $f \colon \mathbb{C} \to \mathbb{C}$ be holomorphic and definable. Then $f$ is a polynomial.
Proof.
If $f$ is not a polynomial, then $f$ has an essential singularity at $\infty$, and so Picard tell us that it takes all but one value infinitely often. Now the inverse image of any element has to be a discrete set in $\mathbb{C} \cong \mathbb{R}^2$, but it also definable, so it has to finite.
Proposition 11. If $f \colon \mathbb{C}^n \to \mathbb{C}$ is holomorphic and definable, then $f$ is a polynomial.
Proof.
Let us write $f$ as $f(\vec{z}, w)$ where the $\vec{z}$ has $n-1$ variables, and expand it as $f(\vec{z}, w) = \sum_{k=0}^\infty a_k(\vec{z}) w^k$. Then $a_k(\vec{z})$ are definable because they can be extracted as derivatives, so by induction they are polynomials. If infinitely many of them were not zero, then there is a $\vec{a}$ such that $a_k(\vec{a}) \neq 0$ for infinitely many $k$, and then $f(\vec{a}, w)$ is not a polynomial in $w$.