Recall that in ACF, Morley rank, Krull dimension, and Lascar rank all agree, and dimension is definable.
Lemma. Let be a Zariski closed subset of , with , with definable over some set . Let be a generic point in , generic over . Then (obviously). Let be the projective space of lines in passing through . Then there is a natural projection map . The image is a Zariski closed subset of (because was proper/complete).
We need to show that if we fix , choose generic in , and then choose generic in , that the line through and intersects only at . By symmetry of forking, we can instead choose first, and then choose generic over and . If we make the choices in this order, then is a generic line through . In particular, , where denotes the code for the set . Suppose intersects at some other point . Then , so
If is irreducible, then is irreducible, on general grounds. Conversely, suppose is not irreducible but is irreducible. Let be the union of and the codes for the irreducible components of . Then and have the same algebraic closure, and is still generic over . Replacing with , we may assume that each irreducible component of is -definable. Write as the union of its irreducible components , with . Let be a generic point in , generic over . By part (a), the map is injective at each , so and are interdefinable over , for each . Therefore for each .
Consequently, so and have the same dimension. But is irreducible because is, and is irreducible by assumption. Therefore .
Meanwhile, , so is strictly smaller than . Consequently . Since , we have . However so for some , contradicting injectivity of at .
Theorem. Let be a formula such that for every , the set is a Zariski closed subset of . Then the set of such that is irreducible is definable.
Proof. We proceed by induction on , the case being extremely easy. Let . Since Morley rank is definable in ACF, the set of such that is definable, as is the set of such that . Breaking into cases, we may assume one of the following:
In the first case, is always irreducible (whenever it is non-empty). In the second case, is irreducible if and only if it is non-empty and the zero set of some irreducible homogeneous polynomial. The set of irreducible homogeneous polynomials of degree is definable, for each , so the set of such that is irreducible is ind-definable. So is the set of such that is not irreducible. (For each , the family of all Zariski closed subsets of which are cut out by at most polynomials of degree at most is a uniformly definable family. The family of all Zariski closed sets of the form with and is also a uniformly definable family. But is the collection of all reducible Zariski closed sets.) Consequently, the set of such that is reducible is definable.
In the third case, proceed as follows. For , let be the projection to with center at . If is generic over , then is irreducible if and only if is. By induction, the set of such that is irreducible is definable. By definability of types in stable theories, the set of such that is irreducible for generic is definable. But this is the set of such that is irreducible. QED
It is easy to prove by induction on that if is a definable (= constructible) subset of , then is a finite union of sets of the form where is an irreducible Zariski closed set and is a Zariski open set intersecting . The Zariski closure of is exactly . (If it were , then , so . Since and , either or . In the first case, we are done; in the second does not intersect .) Writing it follows that
Let be the definable family of all constructible sets of the form where , and each and is cut out by at most polynomials of degree at most , and is irreducible. By the Theorem, is a uniformly definable family. By the previous paragraph, the map assigning to an element of its Zariski closure is also definable.
Now every constructible set is in for sufficiently large . So if is a formula, then compactness ensures that there is some such that for every , . It follows that the Zariski closure of is uniformly definable from . In other words,
Corollary. Zariski closure is definable in families. That is, if for every , then there is some formula such that is the Zariski closure of for every .
From this, we conclude that: