Model theory: algebraically closed fields and compact complex manifolds

Posted on August 28, 2012 by Dima

(this post is from an old blog, covering basics of model-theoretic dimension, and quantifier elimination for two most basic structures studied in geometric model theory: algebraically closed fields and compact complex manifolds. If I wrote this post today I would have chosen a different notation, and would have chosen a different perspective. )

I take a pragmatic stance and fix some “monster model” $\operatorname{\mathcal{M}}$ which is simply a $2^{2^{\aleph_0}}$-saturated model, say (should be enough for doing model theory in any structure related to geometry, it seems); definable sets are subsets of $\operatorname{\mathcal{M}}$ (they are denoted with blackboard bold font: $\operatorname{\mathbb{D}}, \operatorname{\mathbb{E}}, \operatorname{\mathbb{F}}, \operatorname{\mathbb{G}}, \ldots$). I also write $[\varphi]$ for the set of points defined by $\varphi$ in $\operatorname{\mathcal{M}}$, and $[\varphi]_a$ for $\varphi(\operatorname{\mathcal{M}},a)$.

$\omega$-stable theories

Definition. (Morley rank) Let $\varphi$ be a definable set and let $M$ be a model that contains the parameters of $\varphi$. The Morley rank of $\varphi$ with respect $M$ is defined inductively defined as follows:

$\operatorname{MR}^M(\varphi) \geq 0$ if $\varphi(M) \neq \emptyset$;
$\operatorname{MR}^M(\varphi) \geq \alpha + 1$ if there exists infinitely many formulas $\psi_i, [\psi] \subset [\varphi]$ defined over $M$ such that $[\psi_i] \cap [\psi_j] = \emptyset$ and such that $\operatorname{MR}^M(\psi_i)\geq \alpha$;
$\operatorname{MR}^M(\varphi)\geq \alpha$ for a limit ordinal $\alpha$, if there for any $\beta < \alpha$ there exists definable set $\psi, [\psi] \subset [\varphi]$ such that $\operatorname{MR}^M(\psi) \geq \beta$;

Finally, $\operatorname{MR}^M(\varphi) = \alpha$ if $\operatorname{MR}^M(\varphi) \geq \alpha$ but not $\operatorname{MR}^M(\varphi) \geq \alpha + 1$. Morley rank of a definable set is not defined if $\operatorname{MR}^M(\varphi) \geq \alpha$ for any ordinal.

If $M=\operatorname{\mathcal{M}}$, then one omits “with respect to $\operatorname{\mathcal{M}}$” and talks just about “Morley rank”, denoted $\operatorname{MR}(\varphi)$.

Let $\operatorname{\mathbb{D}}$ be a definable set of Morley rank $\alpha$. There exists the maximal number of disjoint sets of Morley rank $\alpha$ into which $\operatorname{\mathbb{D}}$ can be decomposed (if there is no bound, one can construct an infinite tree of nested definable subsets of rank $\alpha$, effectivly contradicting that the Morley rank of $\operatorname{\mathbb{D}}$ is $\alpha$; I omit the details). It is called the of $\operatorname{\mathbb{D}}$ .

Definition. Given a type $p \in S_n(A)$ suppose that there exists a formula $\varphi \in p$ with the minimal Morley rank and degree in $p$ (more precisely, the minimum is attained in the lexicographic order $(\operatorname{MR}(\psi), \operatorname{dM}(\psi))$ for $\psi \in p$). The Morley rank and degree of the type $p$ are defined to be the Morley rank and degree of this formula, or undefined otherwise.

The following lemmas show that if a model $M$ is $\aleph_0$-saturated then $\operatorname{MR}^M$ equals the Morley rank.

Lemma. Let $M$ be an $\aleph_0$-saturated model. Then for any $a,b\in M$ such that $\operatorname{tp}(a/M)=\operatorname{tp}(b/M)$, and any formula $\varphi(x,y)$, $\operatorname{MR}^M(\varphi(x,a))=\operatorname{MR}^M(\varphi(x,b))$.

Proof. Induction on $\operatorname{MR}^M(\varphi(x,a))$. Thee claim of the induction step consists of two points:

$\operatorname{MR}^M(\varphi(x,a))=\operatorname{MR}^M(\varphi(x,b))$ for any $b$ such that $\operatorname{tp}(a/\emptyset)=\operatorname{tp}(b/\emptyset)$;
$\operatorname{MR}^M(\varphi(x,a))=\operatorname{MR}(\varphi(x,a))$.

Let us look at the successor ordinal case. Let $[\psi_i(x,c_i)]$ be definable subsets of $[\varphi(x,a)]$ which is of rank $\geq \alpha+1$. Since $M$ is $\aleph_0$-saturated one can find a sequence of tuples $\{d_i\}$ in $M$ such that \[ \operatorname{tp}(a, c_0, \ldots, c_n/\emptyset)=\operatorname{tp}(b,d_0, \ldots, d_n/\emptyset) \] for any $n$. But then $[\psi_i(M,d_i)]$ are pairwise disjoint and have the same rank as $[\psi_i(M,c_i)]$ by induction hypothesis, and they also are subsets of $\varphi(M,b)$. Therefore $\operatorname{MR}^M(\varphi(x,b)) \geq \alpha+1$.

Lemma 1. Let $M$ be an $\aleph_0$-saturated model and let $\varphi$ be a formula defined over $M$. Then $\operatorname{MR}(\varphi)=\operatorname{MR}^M(\varphi)$.

Proof. Let $N$ be some elementary extension of $N$. Show that $\operatorname{MR}^N(\varphi(x,a))=\operatorname{MR}^M(\varphi(x,a))$ by induction on $\operatorname{MR}^N(\varphi(x,a))$.

Consider subsets $\psi_i(x,c_i)$ with $c_i \in N$ of rank $\alpha$ with respect to $N$. Using $\aleph_0$-saturatedness of $M$ construct a sequence $\{d_i\}$ such that \[ \operatorname{tp}(a, c_0, \ldots, c_n/\emptyset)=\operatorname{tp}(a,d_0, \ldots, d_n/\emptyset) \] for all $n$. Then $[\psi_i(x,c_i)]$ are disjoint and subsets of $[\varphi(x,a)]$, and by induction hypothesis $\operatorname{MR}^M(\psi_i(x,c_i)) \geq \alpha$, hence $\operatorname{MR}^M(\varphi(x, a)) \geq \aleph+1$.

Corollary 2. Let $M$ be an $\aleph_0$-saturated model and let $\varphi$ be a formula defined over $M$. Then there exists a decomposition of $\varphi$ into $\operatorname{dM}(\varphi)$ sets defined over $M$.

Proof. The same as in Lemma 1.

Definition. ($\omega$-stable theory) A theory is called $\omega$-stable if for every $n$-type its Morley rank (not necessarily finite) is defined.

Proposition. In an $\omega$-stable theory any type $p$ is of the form \[ p = \{ \psi(x) \mid \operatorname{MR}([\varphi] \setminus [\psi]) < \operatorname{MR}(p) \} \] where $\varphi$ is a definable set in $p$. It is not uniquely determined: if $\chi$ is such that $\operatorname{MR}([\varphi] \operatorname{\triangle}[\chi]) < \operatorname{MR}(p)$, then $p$ can be equivalently represented as \[ p = \{ \psi(x) \mid \operatorname{MR}([\chi] \setminus [\psi]) < \operatorname{MR}(p) \} \]

Definition. A rank is a function from types to ordinals. A rank $\mathrm R$ is called if for any type $p \in S_n(A), R(p)=\alpha$ there exists $\psi$ such that $\psi \in q$ if and only if $R(q) \leq \alpha$.

Remark. It is an immediate consequence of the definition of Morley rank for types that it is continuous (without any assumptions on the theory).

So intuitively model theory of $\omega$-stable theories is very similar to algebraic geometry in the spirit of Weil: every definable set (“variety”) has a dimension and a generic point of an irreducible variety is a point that does not belong to all definable subsets (“subvarieties”) of lower dimension.

The principal tool of stability theory is definability of types. One needs some tricky combinatorics to deduce it for general stable theories. In the $\omega$-stable case everything boils down to a certain property of Morley rank.

Lemma 3. Let $M$ be an $\aleph_0$-saturated model. Let $[\varphi]$ be a definable set over $M$ of Morley rank $\alpha$ and let $\psi$ be a definable set such that $\operatorname{MR}([\varphi \land \psi])=\alpha$. Then there exists a tuple $a \in M$ such that $a \in [\psi]$.

Proof. Induction on the rank of $\varphi$. If rank is 0 then the intersection $[\varphi] \cap [\psi]$ is just a finite subset of $M$.

Suppose the statement holds for sets of rank $< \alpha$. Let $\varphi$ be of rank $\alpha$. We may assume that $\varphi$ has Morley degree 1, otherwise consider a degree 1 component that has intersection with $[\psi]$ of rank $\alpha$ (it is definable over $M$ by Corollary 2).

Let $\operatorname{MR}([\varphi] \setminus [\psi]) = \beta < \alpha$. By Lemma 1 there exist infinitely many definable over $M$ subsets $[\chi_i(x,c_i)] \subset [\varphi]$ of rank $\beta$. Then $\operatorname{MR}([\chi_i] \setminus [\psi])$ must have rank $< \beta$, except may be finitely many of them. Therefore there exists a set $[\chi_i]$ such that $\operatorname{MR}([\chi_i] \setminus [\psi]) = \beta$.

By induction hypothesis $[\chi_i(x,c_i)] \cap [\psi]$ contains a tuple from $M$.

Lemma 4. Let $\cup_{i \in I} [\phi_i] = \operatorname{\mathcal{M}}\setminus ( \cup_{j \in J} [\psi_j])$. Then there exists a finite set of indices $I_0$, such that $\cup_{i \in I_0} [\varphi_i] = \cup_{i \in I} [\varphi_i]$.

Proof. By compactness, since the infinite intersection $\cap [\neg \phi_i] \cap (\cap [\neg \psi_j])$ is empty, hence there is a finite sub-intersection which is empty, $\cap_{i \in I_0} [\neg \phi_i] \cap (\cap_{j \in J_0} [\neg \psi_j])$, hence $\cap_{j \in J_0} [\neg \psi_j] \subset \cap_{i \in I_0} [\neg \phi_i]$. Therefore,
\[ \bigcup [\phi_i] = \operatorname{\mathcal{M}}\setminus \bigcup [\psi_j] \subset \operatorname{\mathcal{M}}\setminus \bigcup_{j \in J_0} [\psi_j] \subset \bigcap_{j \in J_0} [\neg \psi_j] \subset \bigcap_{i \in I_0} [\neg \phi_i] \]

Theorem 5. (Definability of types) Let $\varphi$ be a definable set in an $\omega$-stable theory. Then for any formula $\psi(x,y)$ the set \[ \{ b \in \operatorname{\mathcal{M}}\mid \operatorname{MR}([\varphi] \setminus [\psi]_b) < \operatorname{MR}([\varphi]) \} \] is definable. We will further denote this set $d_px \psi(x,y)$.

Proof. First, we will show that for any $\psi(x,c)$ there exists a finite sequence of tuples $\{a_i^c\} \subset M$ such that ${a_i^c} _c $ and $\{a_i^c\} \in [\psi]_b$ implies $\operatorname{MR}([\varphi] \setminus [\psi]_b) < \operatorname{MR}([\varphi])$ (but not vice versa!).

We build $\{a_i^c\}$ inductively along with another sequence $\{b_i\}$. Take $a_0 \in M$ to be some point in $M$ (this exists by Lemma 3) such that $a_0 \in [\varphi] \cap [\psi]_c$. If there does not exist $b$ such that $a_0 \notin [\psi]_b$ and $\operatorname{MR}([\varphi] \setminus [\psi]_b) < \operatorname{MR}([\varphi])$, we are finished. Otherwise, pick $a_1 \in [\varphi] \cap [\psi]_c \setminus [\psi]_{b_0}$. Repeat this process inductively. Since $a_i \notin [\psi]_{b_j}$ for $j < i$, the formula $\psi$ has the property: $\psi(a_i, b_j)$ if and only if $i < j$. If the process continues indefinitely, we would have constructed an infinite sequence of tuples $c_i=a_ib_i$, linearly ordered by a definable relation.

A standard argument shows this is not possible in $\omega$-stable theory. By compactness, we can in fact construct a sequence $\{d_r\}_{r\in \operatorname{\mathbb{Q}}}$ of tuples ordered by definable relation $\psi(x,y)$ such that its order type is that of $(\operatorname{\mathbb{Q}}, <)$.Consider the (possibly incomplete) type $p$ consisting of formulas $\chi(x)$ such that $\chi \cap \{d_r\} \supset [0,1]$. Then $p$ has no Morley rank. Indeed, suppose that there exists a formula $\varphi$ of minimal rank and degree in $p$. But then $\varphi \land \neg \psi(x, t)$ and $\varphi \land \psi(x,t)$ for some $t, 0 < t < 1$, either are of the same rank as $\varphi$ but of lower degree, or one of them has a lower rank, which contradicts the choice of $\psi$.

Now let $\theta_c(y)$ be the formula $\land \psi(a^c_i,y)$ — note that it is defined over $M$. Then a conjunction of all formulas of this form $\cup [\theta_c(y)]$ is equal to $[d_px \varphi(x,y)]$. Note that we are not taking union over all $c$, just the formulas that arise when we pick various $c$-s (this is why it is crucial that $\theta_c$ is defined over $M$ — the conjucnction therefore depends only on $M$, not on all possible values of parameter $c$). We need to show that a finite conjunction of $\theta_c$-s suffices.

Repeat all the preivous arguments replacing $\psi$ by $\neg \psi$ and get a set of formulas $\theta'_c(y)$. Then \[ \cup [\theta_c] = \operatorname{\mathcal{M}}\setminus \cup [\theta'_c] \] and hence by Lemma 4 $[d_px \varphi(x,y)]$ is a finite conjunciton of some $\theta_c$.

Corollary. If $p \in S(A)$ then $d_p x \psi(x,y)$ is defined over $A$.

Proof. The definable set $d_p x \psi(x,y)$ is invariant under $\operatorname{Aut}(\operatorname{\mathcal{M}}/A)$: indeed, the Morley rank of $[\varphi] \cap [\psi]_b$ is the same for $b$-s which have the same type, as we have shown while proving Lemma 1.

So we get a definition over $A$ at the cost of losing constructibility: we have no idea what the “honest” definition over $A$ looks like.

Forking in $\omega$-stable theories

Definition. Let $p \in S_n(B)$. One says that $p$ if $\operatorname{MR}(p \restriction A) > \operatorname{MR}(p)$. If $a$ is a tuple such that the type $\operatorname{tp}(a/bA)$ doesn’t fork over $A$, then one says that $a$ is , notation: $a \operatorname{\perp}_A b$ (the lower bar should be round but I cannot represent it using MathJax or HTML entities).

Theorem 6. Suppose $p \in S(A)$ and $\operatorname{dM}(p)=1$. Then for any $B \supset A$ the type \[ q = \{\psi(x,b) \mid b \in B \cap [d_px\psi(x,y)]\} \] is the unique extension of $p$ over $B$ that doesn’t fork over $A$ (non-forking extension). Clearly, $q$ has the same definition as $p$, i.e. is defined over $A$.

Proof. It is clear from the definition of forking that the non-forking extension of $p$ to $B$ is \[ \{\psi(x,b) \mid b \in B, \operatorname{MR}([\varphi] \setminus [\psi(x,b)]) < \operatorname{MR}([\varphi]\} \] and by Theorem 5 this is precisely the type in the statement of the theorem.

Theorem. (Symmetry of forking) For any tuples $a,b$ and any set of parameters $A$, $a \operatorname{\perp}_A b$ if and only if $b \operatorname{\perp}_A a$.

It has been useful for me to realise that the only essential tools that one has in $\omega$-stable theories are definability of types (a fact true in general stable theories as well) and Lemma 3 (a fact particular to $\omega$-stable theories). It is not surprising then that in $\omega$-stable theories one always passes from sets of parameters to models.

Proof. First, suppose $A$ is a actually a model $M$, $\aleph_0$-saturated. Let $\varphi(x)$ be a formula in $\operatorname{tp}(a/M)$ such that $\operatorname{MR}(\varphi)=\operatorname{MR}(a/M)$ and $\psi(y)$ be a formula in $\operatorname{tp}(b/M)$ such that $\operatorname{MR}(\psi)=\operatorname{MR}(b/M)$.

It is actually more convenient to prove that $a \operatorname{\not\perp}_M b$ implies $b \operatorname{\not\perp}_M a$. So suppose that there is a formula $\chi$, such that $\models \chi(a,b)$ and $\operatorname{MR}(\chi(\operatorname{\mathcal{M}},b)) < \operatorname{MR}(\varphi)$. Consider the formula \[ \theta(y)=\psi(y) \land \chi(a, y) \land \neg d_{tp(a/M)}x\,(\varphi(x) \land \chi(x,y)) \] Suppose that $b \operatorname{\perp}_M a$, so $\operatorname{MR}(\theta)=\operatorname{MR}(\psi)$ (since $\theta$ is defined over $Mb$). Then $\operatorname{MR}( \neg d_{tp(a/M)}x\,(\varphi(x) \land \chi(x,y) \land \psi(y))=\operatorname{MR}(\psi(y))$ and by Lemma 3 there exists a tuple $m \in M$ such that $m \in [\operatorname{MR}( \neg d_{tp(a/M)}x\,(\varphi(x) \land \chi(x,y) \land \psi(y)]$, or in other words, $\operatorname{MR}(\varphi(x) \land \chi(x,m) < \operatorname{MR}(\varphi)$ which is the contradiction to the minimality of the rank of $\varphi$ in $\operatorname{tp}(a/M)$.

To prove the general case we will show that for a model $M$ containing the parameter set $A$ there exist $a', b'$ such that $a' \operatorname{\perp}_M b'$ if and only if $a \operatorname{\perp}_A b$. This is just several consecutive applications of Theorem 6.

Let $b'$ be a realisation of a nonforking extension of $\operatorname{tp}(b/A)$ to $M$. Let $c$ be a tuple such that $\operatorname{tp}(cb'/)=\operatorname{tp}(ab/A)$ and let $a'$ be a realisation of a non-forking extension of $\operatorname{tp}(c/Ab')$ to $Mb'$. Then \[ \operatorname{MR}(a'/Mb')=\operatorname{MR}(c/Ab')=\operatorname{MR}(a/Ab), \operatorname{MR}(a'/M)=\operatorname{MR}(a/A) \textrm{ and } \operatorname{MR}(b'/M)=\operatorname{MR}(b) \] and the conclusion follows.

Strongly minimal sets

A strongly minimal set $\operatorname{\mathbb{D}}$ is a definable set that has Morley rank and Morley degree 1.

Given a strongly minimal set $[\varphi]$ and a model $M$ one can consider the structure with the universe $\varphi(M)$ and definable sets traces of the definable sets of the ambient theory on $\varphi(M)$.

Proposition. Let $[\varphi]$ be definable set in a stable theory. Then for any $\psi(x,c)$ there exists a $c' \in [\varphi]$ and $\psi'(x,y)$ such that $[\varphi] \cap [\psi(x,c)] = [\varphi] \cap [\psi'(x,c')]$. (This property of $[\varphi]$ is called ).

Proof. The formula $\varphi(a) \land \psi(a,y)$ belongs to $\operatorname{tp}(c/[\varphi])$ for any $a \in [\varphi(x) \land \psi(x,c)]$ so $\varphi(x) \land \psi(x,c) = d_{\operatorname{tp}(c/[\varphi])}y\, \psi(x,y)$ which is definable over $[\varphi]$.

Definition. (Algebraic dimension) A tuple $a$ of elements of a model $M$ is called if for any $i$, $a_i \notin \operatorname{acl}(a_{\neq i}B)$. A tuple $a$ is said to have $n$ over $B$ if any indpendent subtuple $b \subset a$ is of size $\leq n$.

The , denoted $\dim([\varphi])$, is the supremum of algebraic dimensions of tuples $a \in M$ such that $M \models \varphi(a)$.

In fact, algebraic dimension behaves best when $\operatorname{acl}$ satisfies the exchange property:

If $a \in \operatorname{acl}(b \cup A) \setminus \operatorname{acl}(A)$ then $b \in \operatorname{acl}(a \cap A)$.

In this case the maximal independent subtuples of a set are of the same cardinality. Indeed, if $c_I, c_J$ are maximal independent subtuples of $c$ then for any $i \in I$ there exists some $j \in J$ such that $c_j (c_I c_i) $ and by exchange property $c_i \in \operatorname{acl}(c_I \setminus c_i \cup c_j)$, hence $c_I \setminus c_i \cup c_j$ is a maximal independent subtuple. If $|I| < |J|$ then continuing like this we will find a proper subtuple of $c_J$ which is maximal independent, coming to contradiction.

Proposition If $T$ is strongly minimal then $\operatorname{acl}$ satisfies the exchange property.

Proof. This is a consequence of symmetry of forking since $b \notin acl(a)$ if and only if $\operatorname{tp}(a/Ab)$ doesn’t fork over $A$.

Note that in order to prove symmetry of forking in this particular case (for elements of a model, not tuples) we only need definability of 1-types, which in the strongly minimal case is straightforward: for non-algebraic type $p \in S_1(A)$ we only have to show that the set $\{b \in \operatorname{\mathcal{M}}\mid [\psi]_b \textrm{ is infinite }\}$ is definable for any formula $\psi$. But this is easy, since there exists a number $n$ such that $|[\psi]_b| > n$ implies $[\psi]_b$ is infinite — if it wouldn’t exist, by compactness one could find $c$ such that $[\varphi]_c$ and its complement are infinite, contradicting strongly minimality.

This property of a theory has a name.

Definition. (Elimination of $\exists^\infty$) A theory $T$ is said to if for every formula $\varphi(x,y)$ the set of parameters $a$ in some model $M$, such that $\varphi(M,a)$ is infinite, is definable.

So what we have just shown is

Proposition. A strongly minimal theory eliminates $\exists^\infty$.

Lemma 7. In an $\omega$-stable theory, let $\bar{b} \in \operatorname{acl}(a_1,\ldots,a_n,A)$. Then $\operatorname{MR}(\bar{b}/A) \leq \operatorname{MR}(\operatorname{tp}(a_1, \ldots, a_n/A))$.

Proof. Suppose that $\varphi$ is the definable set of minimal rank in $\operatorname{tp}(\bar a)$ and that $\psi(x,y)$ is the definable set of minimal rank in $\operatorname{tp}(\bar a, b)$.

Clearly, $\operatorname{MR}(\bar a, b) \geq \operatorname{MR}(\bar a)$: by induction on the rank of $\varphi$, any infinite family $\chi_i(x)$ of disjoint subsets of $\varphi(x)$ of rank $\alpha$ gives rise to a family of disjoint subsets $\chi_i(x) \land \psi(x,y)$ which is of rank $\alpha$ by induction hypothesis.

Now we will prove by induction on $\operatorname{MR}(\psi)$ that $\operatorname{MR}(\psi) \geq \alpha$ implies $\operatorname{MR}(\varphi) \geq \alpha$. We may assume without loss of generality that for any $c \in [\varphi]$, the fibre $[\psi]_c$ has exactly $n$ elements, where $n=[\psi]_a$ is the number of conjugates of $b$ over $a$. Indeed, the formula $\varphi(x) \land |[\psi]_x|=n$ belongs to $\operatorname{tp}(\bar a)$ so is of the rank $\geq \operatorname{MR}(\varphi)$, similarly $\psi(x,y) \land |[\psi]_x|=n$ belongs to $\operatorname{tp}(\bar a,b)$.

Suppose $\operatorname{MR}(\psi)=\alpha+1$ and suppose for the sake of contradiction that $\operatorname{MR}(\varphi)=\alpha$. So there exist definable sets $\chi_i(x,y)$, $[\chi_i] \subset [\psi]$ of rank $\alpha$. Then by induction hypothesis $\exists y\, \chi_i(x,y)$ are of rank $\alpha$.

There are finitely many — $\operatorname{dM}(\varphi)$ — rank $\alpha$ components of $\varphi$, so there must by at least one of them (of Morley degree one), $\varphi_0$ such that $\exists y\, \chi_i(x,y)$ has intersection of rank $\alpha$ with it for infinitely many $i$, say for indices lying in $I_0$. By compactness there is a point $c \in \cap_{i \in I_0} [\exists y\, \chi_i(x,y)] \cap [\varphi_0]$. But then, since $\chi_i$ are all distinct, $[\psi]_c$ must have infinitely many points, a contradiction.

Lemma. Let $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ be two sets of mutually independent points (over $A$, say). Then $\operatorname{tp}(a_1, \ldots, a_n/A) = \operatorname{tp}(b_1,\ldots, b_n/A)$.

Proof. For notational simplicity suppose $A=\emptyset$.

Let $n=1$. Then any $\varphi \in \operatorname{tp}(a)$ is of the form $\operatorname{\mathcal{M}}\setminus \{c_1, \ldots, c_n\}$ where $\{c_1, \ldots, c_n\} \subset \operatorname{acl}(\emptyset)$ and since $b \notin \operatorname{acl}(\emptyset)$, $b \in [\varphi]$.

Suppose the statement is true for $n$. Let $a_1, \ldots, a_n, a_{n+1}$ and $b_1, \ldots, b_n, b_{n+1}$ be independent sets. For any $\varphi \in \operatorname{tp}(\bar a)$, $[\varphi(x, a_1, \ldots, a_n)]$ is of the form $\operatorname{\mathcal{M}}\setminus \bar c$ where $\bar c \subset \operatorname{acl}(a_1, \ldots, a_n)$. The formula $(\exists^{=n}x \neg \varphi(x, y_1, \ldots, y_n)$ belongs to $\operatorname{tp}(a_1, \ldots, a_n)$ and hence to $\operatorname{tp}(b_1, \ldots, b_n)$ by induction hypothesis. Since $b_{n+1} \notin \operatorname{acl}(b_1, \ldots, b_n)$, $b_{n+1} \in [\varphi(x,b_1, \ldots, b_n)]$, qfd.

Lemma 8. In a strongly minimal theory for any definable set $\operatorname{\mathbb{D}}$, $\dim(\operatorname{\mathbb{D}})=\operatorname{MR}(\operatorname{\mathbb{D}})$.

Proof. Let $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ be two sets of mutually independent points (over $A$, say). We will prove that $\operatorname{MR}(\operatorname{tp}(a_1, \ldots, a_n/A))=n$. Together with Lemma 7 this would imply the desired.

For any non-algebraic element $a$ in a strongly minimal theory the type $\operatorname{tp}(a/\emptyset)$ is just $\{ x \neq a \mid a \in \operatorname{acl}(\emptyset)\}$ and the rank of this type is clearly 1.

Suppose that the statement is proved for $n$. Let $a_1, \ldots, a_{n+1}$ be independent and let $\varphi$ be the set with the minimal Morley rank and degree in $\operatorname{tp}(\bar a)$. The formula $\exists x_2,\ldots,x_{n+1} \varphi$ must have rank 1 since $\bar a$ is independent. So there exist infinitely many elements $\{b_i\}$ such that $x_1=b_i$ is consistent with $\varphi(\bar x)$.

Let $c_2, \ldots, c_n$ be some tuple of elements independent over $\{b_i\}$. Then by Lemma 8, $\operatorname{tp}(b_i, \bar c)=\operatorname{tp}(\bar a)$ and $(c_2, \ldots, c_n) \in [\exists x_1 \varphi(x_1, x_2, \ldots,x_n) \land x_1=b_i]$ and the rank of the latter formula is $n-1$ by induction hypothesis. Hence there are infinitely many rank $\geq n-1$ definable subsets of $\varphi$ and $\operatorname{MR}(\varphi)\geq n$.

Now let us show that $\operatorname{MR}(\varphi) \leq n$. In fact it is also true that $\operatorname{dM}(\varphi)=1$; we will show this by proving that for any formula $[\psi] \subset [\varphi]$ such that $\bar a \notin [\psi]$, $\operatorname{MR}(\psi) < n$. Indeed, for any tuple $\bar b \in [\psi]$ if $\bar b=(b_1, \ldots, b_n)$ were independent then $\operatorname{tp}(\bar b)=\operatorname{tp}(\bar a)$ which would contradict the fact that $\bar a \notin [\psi]$. Therefore, there exists a maximal number $l < n$ such that $(b_1, \ldots, b_l)$ are independent. By induction hypothesis and Lemma 7, $\operatorname{MR}(\bar b)=l$ hence $\operatorname{MR}(\psi) \leq l < n$.

Proposition. If a theory eliminates $\exists^\infty$ then algebraic dimension is definable, i.e. for any formula $\psi(x,y)$ the set \[ \{ b \in \operatorname{\mathcal{M}}\mid \dim [\psi]_b \geq l\} \] is definable for any $l$.

Proof. Induction on the number $n$ of variables in the tuple $x$.

For $n=1$, by there are only two possibilities: the fibre $[\psi]b$ is 0-dimensional (finite) or 1-dimensional. Definability of dimension is equivalent to elimination of $\exists^\infty$ in this case.

Suppose the statement is true for some $n$. Let $\psi(x_1, \ldots, x_{n+1}, y)$ be a formula. Induction on $l$. The formula that defines 0-dimensional fibres is again obtained from elimination of $\exists^\infty$. The set \[ \{ b \in \operatorname{\mathcal{M}}\mid \dim [\psi]_b \geq l+1 \} \] is the union of two sets \[ \begin{array}{c} \{ b \in \operatorname{\mathcal{M}}\mid \dim [\exists x_{n+1} \psi(x_1, \ldots, x_{n+1}, b)] \geq l+1 \} \\ \qquad \textrm{ and } \qquad \\ \{ b \in \operatorname{\mathcal{M}}\mid \dim [\exists^\infty x_{n+1} \psi(x_1, \ldots, x_{n+1}, b)] \geq l \} \\ \end{array} \] Indeed, by dimension formula $\dim(\bar c, d)=\dim(\bar c) + \dim(d/\bar c)$ and for subsets of $\operatorname{\mathcal{M}}$, $\sup \dim_{a\in [\chi]} = 1$ if and only if $[\chi]$ is infinite. But then the desired set is definable by induction hypothesis and elimination of $\exists^\infty$.

Corollary. In a strongly minimal theory Morley rank is definable.

An is one where every model is in the $\operatorname{acl}$ of an $\emptyset$-definable strongly minimal set. Note that almost strongly minimal set eliminates $\exists^\infty$, but in general one no longer has $\dim=\operatorname{MR}$.

Examples: algebraically closed fields and compact complex manifolds

The theory of algebraically closed fields of fixed characteristic is strongly minimal. It eliminates quantifiers in the field language (this is a standard result, easily proved using back-and-forth technique). It is clear from the definitions that definable sets in $\operatorname{ACF}_p$ are constructible sets, boolean combinations of closed subsets of $\operatorname{\mathbb{A}}^n$.

In fact the elimination of quantifiers is a particular case of Chevalley’s theorem that works in the general framework of schemes.

Chevalley’s theorem. Let $f: X \to Y$ be a morphism of Noetherian schemes and let $f$ be of finite type. Then for any constructible subset $Z \subset X$ the image $f(Z)$ is constructible.

Proof. We might as well restrict to the closure of $f(Z)$, so we suppose $f$ dominant. Since being constructible is a local property, we may cover $Z$ with open affines and prove the theorem for each open, so we may assume that $Z$ is affine.

Since $Z$ is of the form $Z = \cup Z_i \setminus S_i$ where $Z_i$ are closed irreducible and union of images is the image of unions, suffices to show that each $f(Z_i \setminus S_i)$ is constructible. In fact, it suffices to show that $f(Z_i \setminus S_i)$ contains a set $U$ relatively open in $Y^0=\overline{f(Z_i \setminus S_i)}$ and then use Noetherian induction. For that we may also assume that $S_i$ is defined by a single element of the ring of functions of $Z_i$. But then $Z_i \setminus S_i$ is affine, so we might as well prove the statement for an affine $Z_i'=Z_i \setminus S_i$

Since $f$ is of finite type, $Z_i' \subset \operatorname{\mathbb{A}}_Y^n$. Let $Z_i=\operatorname{Spec}B$ and $Y'=\operatorname{Spec}A, A \subset B$.

Suppose $n=1$. There are two possibilities: $B=A[t]$ or $B=A[t]/I$ for some non-trivial ideal $I$. In the first case it is not hard to see that the projection is an open mapping. In the second case by generic freeness there is an open subset of $D(a) \subset Y'$ such that $B_a$ is a free module over $A_a$, therefore the projection of $Z_i'$ contains $D(a)$.

For $n>1$ factor the map as a sequence of maps as in the previous paragraph.

The theory $\operatorname{ACF}_p$ also eliminates imaginaries. This can be shown in two steps: $\operatorname{ACF}_p$ eliminates imaginaries up to finite ones because it is strongly minimal and $\operatorname{acl}(\emptyset)$ is infinite (a standard lemma due to Pillay and Ziegler), secondly, existence of finite quotients follows from the fact that quotients of affine varieties by finite group actions exist (also a standard result) and quantifier elimination.

Consider the multi-sorted structure with sorts consisting of points of compact complex spaces (complex analytic analogue of algebraic varieties). The language of the structure consists of all closed subspaces (analytic sets) of all Cartesian powers of spaces, with the natural interpretation (“Zariski language”). Of course, gathering all the spaces in one structure is purely notational convention: any definable formula defines a set in one particular sort, no one usually works with the structure sort by sort.

Lemma (“Global irreducible decomposition”, Grauert-Remmert, Chapter 9, § 2) Let $X$ be a complex space and let $Z$ be an analytic set. Then there exists a decomposition \[ Z = \cup_i Z_i \] where $Z_i$ are analytic sets irreducible in the Zariski topology. If $M$ is compact then there are finitely many $Z_i$ in the decomposition.

Corollary.a The topology on $X$ where closed sets are analytic sets (analytic “Zariski topology”) is Noetherian if $X$ is compact.

Theorem 9. (Semicontinuity of fibre dimension) Let $f: X \to Y$ be a map of complex analytic spaces. Then for any $n$ the set \[ \{x \in X\ \mid \ \dim(X_{f(x)}) > n \} \] is analytic in $X$.

Theorem. (Proper mapping theorem) Let $f: X \to Y$ be a proper map (i.e. preimage of any closed set is compact). Then the image of any analytic subset of $X$ in $Y$ is analytic.

Theorem. Let $M$ be a compact complex space considered as a structure. Then $M$ eliminates quantifiers in the Zariski language.

Proof. Let $f: M^{n+1} \to M^n$ be a projection on first $n$ coordinates. We have to show that for any set of the form $Z \setminus S$ where $Z, C$ are irreducible analytic, $f(Z \setminus C)$ contains an set relatively open in $\overline(f(Z))$. Let $n$ be the minimal dimension of the fibre of the map $f$. The set \[ \{x \in X\ \mid \ \dim(X_{f(x)}) >= n+1 \} \] is analytic by Theorem 9 and hence, by Proper Mapping Theorem, the set \[ \{y \in Y\ \mid \ X_y \neq \emptyset \} = f(X) \] is Zariski open in $Y$.

Compact complex spaces eliminate imaginaries (the argument sketch was proposed by Pillay and detailed out in the thesis of Rahim Moosa; I might make a note about it some time). Note however that individual compact complex spaces not necessarily eliminate imaginaries.

For example, consider an elliptic curve $E$ in the Zariski language. By Noether normalisation theorem it projects onto the projective line with finite fibres $f: E \to \mathbb{P}^1$. Therefore the equivalence relation $\sim$ defined as $a \sim b \Leftrightarrow f(a)=f(b)$ is a definable set in $E \times E$. Suppose that $E/\sim$ is definable in $E^n$ along with a projection $E \to E/\sim$. Consider some embedding $E \hookrightarrow \mathbb{C}^n$ where $\mathbb C$ is an algebraically closed field. The quotient $E/\sim$ is definable in $\mathbb C$ and is $\mathbb{P}^1$, therefore is definably isomorphic to $(E/\sim) \subset E^n$. Consider a coordinate projection of $E/\sim$ with dominant image. By composing it with a definable bijection $\mathbb{P}^1 \to E/\sim$ we obtain a definable map $\mathbb{P}^1 \to E$. By quantifier elimination we can find a dominant bijective regular map from an open subset of $\mathbb{P}^1$ to $E$. Since these are curves, by a standard argument this morphism extends to $\mathbb{P}^1$ and we obtain a map from genus 0 to genus 1 curve which cannot be.

Note that Morley rank coincides with independence-theoretic dimension and hence is definable in ACF, by general argument valid in any strongly minimal structure. In compact complex spaces, however, dimension and so is Morley rank (an example?). However, the complex-analytic dimension is definable (as follows from Theorem 9 and elimination of quantifiers).

Now I will briefly explain the meaning of model-theoretic notions such as non-standard parameters, non-algebraic types and forking in geometric terms.

In ACF or in a compact complex space, every type over the empty set is the generic type of a definable set by the general $\omega$-stability, and therefore it is a generic type of a variety (complex analytic or algebraic).

If $\varphi \subset X$ is defined over $\bar a$ where $\bar a$ is a generic point of a variety $Y$ then $[\varphi] = Z_{\bar a}$ where $Z \subset X \times Y$, i.e. $[\varphi]$ is the generic fibre of $Z$. If one considers another set of parameters $\bar a$ which is a generic point of a variety $Y'$ and such that $\bar{a} \subset \operatorname{dcl}(\bar{b})$ then $[\varphi]$ can be thought of as the generic type of a pull-back family of varieties over $Y'$ \[ \begin{array}{ccccc} Z \times_{Y'} Y & & Z & \subset & Y \times X \\ \downarrow & & \downarrow^\pi & & \downarrow\\ Y' & \xrightarrow{p} & Y & = & Y\\ \end{array} \] Here the fibre product is understood in the set-theoretic sense, i.e. \[ Z \times_{Y'} Y = \{\ (y, z) \in Y' \times Z\ \mid\ p(y)=\pi(z)\ \} \] where $p: Y' \to Y$ is the definable set induced by the inclusion $\bar{a} \subset \operatorname{dcl}(\bar{b})$. In case of algebraically closed fields and compact complex spaces if $Z, Y, Y'$ are varieties and $p,\pi$ morphisms this fibre product coincides with the reduction of the fibre product in the respective category.

Let $p$ be a type over $B=\operatorname{dcl}(b_1, \ldots, b_n)$, where $\bar b$ is the generic point of a variety $Y$. Let $\bar{a} \subset B$ and let $\bar{a}$ be a generic point of a definable set $Y'$. Then $p \restriction \bar{a}$ is the generic type of the generic fibre of a fibration $Z \subset Y' \times X$ such that the generic fibre of $Z' \times_Y Y' \cap Z$ has strictly smaller rank than the generic fibre of $Z$.

Let $C=\operatorname{dcl}(\bar c) \supset B=\operatorname{dcl}(\bar b)$, and let $\bar c$ be the generic point of $Y$, $\bar b$ be the generic point of $X$. Let $p$ be the generic type of a variety $Z$ over $X$. Then the non-forking extension of $p$ to $C$, $p \restriction C$ is a the generic type of the generic fibre of the pull-back $Z \times_X Y$.

Model theory: algebraically closed fields and compact complex manifolds

\(\omega\)-stable theories

Forking in \(\omega\)-stable theories

Strongly minimal sets

Examples: algebraically closed fields and compact complex manifolds