## FANDOM

78 Pages

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.)