Recall that in ACF, Morley rank, Krull dimension, and Lascar rank all agree, and dimension is definable.

Lemma. Let {\displaystyle V} be a Zariski closed subset of {\displaystyle \mathbb {P} ^{n}}, with {\displaystyle \dim V\leq n-2}, with {\displaystyle V} definable over some set {\displaystyle C}. Let {\displaystyle p} be a generic point in {\displaystyle \mathbb {P} ^{n}}, generic over {\displaystyle C}. Then {\displaystyle p\notin V} (obviously). Let {\displaystyle P'} be the projective space of lines in {\displaystyle \mathbb {P} ^{n}} passing through {\displaystyle p}. Then there is a natural projection map {\displaystyle \pi :\mathbb {P} ^{n}\setminus p\twoheadrightarrow P'}. The image {\displaystyle \pi (V)} is a Zariski closed subset of {\displaystyle P'} (because {\displaystyle V} was proper/complete).


If {\displaystyle W} is an irreducible component of {\displaystyle V}, and {\displaystyle q} is a generic point on {\displaystyle W} (generic over {\displaystyle p} and {\displaystyle C}), then the projection {\displaystyle V\to \pi (V)} is injective at {\displaystyle q}, i.e., there is no {\displaystyle q'\in V} with {\displaystyle q'\neq q} but {\displaystyle \pi (q')=\pi (q)}.
{\displaystyle \pi (V)} is irreducible if and only if {\displaystyle V} is.



We need to show that if we fix {\displaystyle V}, choose {\displaystyle p} generic in {\displaystyle \mathbb {P} ^{n}}, and then choose {\displaystyle q} generic in {\displaystyle W}, that the line {\displaystyle L} through {\displaystyle p} and {\displaystyle q} intersects {\displaystyle V} only at {\displaystyle q}. By symmetry of forking, we can instead choose {\displaystyle q} first, and then choose {\displaystyle p} generic over {\displaystyle C} and {\displaystyle q}. If we make the choices in this order, then {\displaystyle L} is a generic line through {\displaystyle q}. In particular, {\displaystyle U([L]/Cq)=n-1}, where {\displaystyle [L]} denotes the code for the set {\displaystyle L}. Suppose {\displaystyle L} intersects {\displaystyle V} at some other point {\displaystyle q'}. Then {\displaystyle [L]\in \operatorname {dcl} (qq')}, so {\displaystyle n-1>U(V)\geq U(q'/Cq)\geq U([L]/Cq)=n-1.}


If {\displaystyle V} is irreducible, then {\displaystyle \pi (V)} is irreducible, on general grounds. Conversely, suppose {\displaystyle V} is not irreducible but {\displaystyle \pi (V)} is irreducible. Let {\displaystyle C'} be the union of {\displaystyle C} and the codes for the irreducible components of {\displaystyle V}. Then {\displaystyle C'} and {\displaystyle C} have the same algebraic closure, {\displaystyle U(p/C')=U(p/C)} and {\displaystyle p} is still generic over {\displaystyle C'}. Replacing {\displaystyle C} with {\displaystyle C'}, we may assume that each irreducible component of {\displaystyle V} is {\displaystyle C}-definable. Write {\displaystyle V} as the union of its irreducible components {\displaystyle W_{1}\cup W_{2}\cup \cdots \cup W_{n}}, with {\displaystyle \dim W_{1}\leq \dim W_{2}\leq \cdots \leq \dim W_{n}=\dim V}. Let {\displaystyle (q_{1},\ldots ,q_{n})} be a generic point in {\displaystyle W_{1}\times \cdots \times W_{n}}, generic over {\displaystyle Cp}. By part (a), the map {\displaystyle V\to \pi (V)} is injective at each {\displaystyle q_{i}}, so {\displaystyle q_{i}} and {\displaystyle \pi (q_{i})} are interdefinable over {\displaystyle Cp}, for each {\displaystyle i}. Therefore {\displaystyle U(\pi (q_{i})/Cp)=U(q_{i}/Cp)} for each {\displaystyle i}.

Consequently, {\displaystyle \dim V\geq \dim \pi (V)\geq \dim \pi (W_{n})\geq U(\pi (q_{n})/Cp)=U(q_{n}/Cp)=\dim W_{n}=\dim V,} so {\displaystyle \pi (W_{n})} and {\displaystyle \pi (V)} have the same dimension. But {\displaystyle \pi (W_{n})} is irreducible because {\displaystyle W_{n}} is, and {\displaystyle \pi (V)} is irreducible by assumption. Therefore {\displaystyle \pi (W_{n})=\pi (V)}.

Meanwhile, {\displaystyle W_{1}\neq W_{n}}, so {\displaystyle W_{1}\cap W_{n}} is strictly smaller than {\displaystyle W_{1}}. Consequently {\displaystyle \dim \pi (W_{1}\cap W_{n})\leq \dim W_{1}\cap W_{n}<\dim W_{1}}. Since {\displaystyle U(\pi (q_{1})/Cp)=U(q_{1}/Cp)=\dim W_{1}>\dim \pi (W_{1}\cap W_{n})}, we have {\displaystyle \pi (q_{1})\notin \pi (W_{1}\cap W_{n})}. However {\displaystyle \pi (q_{1})\in \pi (V)=\pi (W_{n}),} so {\displaystyle \pi (q_{1})=\pi (r)} for some {\displaystyle r\in W_{n}\setminus W_{1}}, contradicting injectivity of {\displaystyle \pi } at {\displaystyle q_{1}}.


Theorem. Let {\displaystyle \phi (x;y)} be a formula such that for every {\displaystyle b}, the set {\displaystyle \phi (\mathbb {U} ;b)} is a Zariski closed subset of {\displaystyle \mathbb {P} ^{n}}. Then the set of {\displaystyle b} such that {\displaystyle \phi (\mathbb {U} ;b)} is irreducible is definable.

Proof. We proceed by induction on {\displaystyle n}, the case {\displaystyle n=1} being extremely easy. Let {\displaystyle V_{b}=\phi (\mathbb {U} ;b)}. Since Morley rank is definable in ACF, the set of {\displaystyle b} such that {\displaystyle \dim V_{b}<n-1} is definable, as is the set of {\displaystyle b} such that {\displaystyle \dim V_{b}=n}. Breaking into cases, we may assume one of the following:

In the first case, {\displaystyle V_{b}} is always irreducible (whenever it is non-empty). In the second case, {\displaystyle V_{b}} 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 {\displaystyle d} is definable, for each {\displaystyle d}, so the set of {\displaystyle b} such that {\displaystyle V_{b}} is irreducible is ind-definable. So is the set of {\displaystyle b} such that {\displaystyle V_{b}} is not irreducible. (For each {\displaystyle m}, the family {\displaystyle {\mathcal {F}}_{n}} of all Zariski closed subsets of {\displaystyle \mathbb {P} ^{n}} which are cut out by at most {\displaystyle m} polynomials of degree at most {\displaystyle m} is a uniformly definable family. The family {\displaystyle {\mathcal {R}}_{n}} of all Zariski closed sets of the form {\displaystyle W_{1}\cup W_{2}} with {\displaystyle W_{1}\not \subseteq W_{2}\not \subseteq W_{1}} and {\displaystyle W_{i}\in {\mathcal {F}}_{n}} is also a uniformly definable family. But {\displaystyle \bigcup _{n=1}^{\infty }{\mathcal {R}}_{n}} is the collection of all reducible Zariski closed sets.) Consequently, the set of {\displaystyle b} such that {\displaystyle V_{b}} is reducible is definable.

In the third case, proceed as follows. For {\displaystyle p\in \mathbb {P} ^{n}}, let {\displaystyle \pi _{p}} be the projection to {\displaystyle \mathbb {P} ^{n-1}} with center at {\displaystyle p}. If {\displaystyle p} is generic over {\displaystyle b}, then {\displaystyle \pi _{p}(V_{b})} is irreducible if and only if {\displaystyle V_{b}} is. By induction, the set of {\displaystyle (p,b)} such that {\displaystyle \pi _{p}(V_{b})} is irreducible is definable. By definability of types in stable theories, the set of {\displaystyle b} such that {\displaystyle \pi _{p}(V_{b})} is irreducible for generic {\displaystyle p} is definable. But this is the set of {\displaystyle b} such that {\displaystyle V_{b}} is irreducible. QED

It is easy to prove by induction on {\displaystyle \dim D} that if {\displaystyle D} is a definable (= constructible) subset of {\displaystyle \mathbb {P} ^{n}}, then {\displaystyle D} is a finite union of sets of the form {\displaystyle V\cap U} where {\displaystyle V} is an irreducible Zariski closed set and {\displaystyle U} is a Zariski open set intersecting {\displaystyle V}. The Zariski closure of {\displaystyle V\cap U} is exactly {\displaystyle V}. (If it were {\displaystyle W}, then {\displaystyle V\cap U\subset W}, so {\displaystyle V\subset W\cup (V\setminus U)}. Since {\displaystyle W\subset V} and {\displaystyle V\setminus U\subset V}, either {\displaystyle W=V} or {\displaystyle V\setminus U=V}. In the first case, we are done; in the second {\displaystyle U} does not intersect {\displaystyle V}.) Writing {\displaystyle D=\bigcup _{i=1}^{m}V_{i}\cap U_{i},} it follows that {\displaystyle {\overline {D}}=\bigcup _{i=1}^{m}V_{i}.}

Let {\displaystyle {\mathcal {F}}_{m}} be the definable family of all constructible sets of the form {\displaystyle \bigcup _{i=1}^{k}V_{i}\cap U_{i}} where {\displaystyle k\leq m}, and each {\displaystyle V_{i}} and {\displaystyle U_{i}^{c}} is cut out by at most {\displaystyle m} polynomials of degree at most {\displaystyle m}, and {\displaystyle V_{i}} is irreducible. By the Theorem, {\displaystyle {\mathcal {F}}_{m}} is a uniformly definable family. By the previous paragraph, the map assigning to an element of {\displaystyle {\mathcal {F}}_{m}} its Zariski closure is also definable.

Now every constructible set is in {\displaystyle {\mathcal {F}}_{m}} for sufficiently large {\displaystyle m}. So if {\displaystyle \phi (x;y)} is a formula, then compactness ensures that there is some {\displaystyle m} such that for every {\displaystyle b}, {\displaystyle \phi (\mathbb {U} ;b)\in {\mathcal {F}}_{m}}. It follows that the Zariski closure of {\displaystyle \phi (\mathbb {U} ;b)} is uniformly definable from {\displaystyle b}. In other words,

Corollary. Zariski closure is definable in families. That is, if {\displaystyle \phi (\mathbb {U} ;b)\subset \mathbb {P} ^{n}} for every {\displaystyle b}, then there is some formula {\displaystyle \psi (x;y)} such that {\displaystyle \psi (\mathbb {U} ;b)} is the Zariski closure of {\displaystyle \phi (\mathbb {U} ;b)} for every {\displaystyle b}.

From this, we conclude that: