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

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

Then,

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

Proof.

(a)

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

(b)

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

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

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

QED

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

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

$\dim V_{b}=n$ whenever $V_{b}$ is non-empty.
$\dim V_{b}=n-1$ whenever $V_{b}$ is non-empty.
$\dim V_{b}<n-1$ whenever $V_{b}$ is non-empty.

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

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

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

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

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

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

From this, we conclude that:

The property of being Zariski closed (in $\mathbb {P} ^{n}$ ) is definable in families.
The property of being Zariski closed in $\mathbb {A} ^{n}$ is definable in families. (Use the embedding of $\mathbb {A} ^{n}$ into $\mathbb {P} ^{n}$ .)
The property of being Zariski closed and irreducible in $\mathbb {A} ^{n}$ is definable in families. (A closed subset of $\mathbb {A} ^{n}$ is irreducible if and only if its Zariski closure in $\mathbb {P} ^{n}$ is irreducible.) ::: ::: :::