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 \le 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' \ne 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) \ge U(q'/Cq) \ge 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 \le \dim W_2 \le \cdots \le \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 \ge \dim \pi(V) \ge \dim \pi(W_n) \ge 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 \ne W_n $, so $ W_1 \cap W_n $ is strictly smaller than $ W_1 $. Consequently $ \dim \pi(W_1 \cap W_n) \le \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 \le 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.)