Proofs:Irreducibility is Definable

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.)