Parameters and strongly minimal sets

Question

Suppose $T$ is a countable complete theory, with monster model $\mathbb{C}$. A definable set $D := \phi(\mathbb{C}, \overline a)$ is strongly minimal if given any other formula $\psi(x, \overline b)$ with $\overline b\in \mathbb{C}$, then $D\cap \psi(\mathbb{C}, \overline b)$ is either finite or cofinite.

One of the important facts about strongly minimal sets is that the closure operator $A\rightarrow acl(A)\cap D$ turns $D$ into a pregeometry. My question is about proving the exchange principle: if $A\subseteq D$; $b, c\in D$; $b\in acl(A\cup \{c\})\setminus acl(A)$, then $c\in acl(A\cup \{b\})$.

Most proofs I find start as follows. Suppose not, and find $b, c$ as above but with $c\not\in acl(A\cup \{b\})$. Now there is a step where we include $A$ and $\overline a$ as constants in the language. Including $A$ is fine, but including $\overline a$ makes me uneasy. It could be the case that $c\in acl(A\cup\{b\} \cup \{\overline a\})$, which would undermine the rest of the proof.

If anyone knows of a good proof of exchange in this setting, I would much appreciate it.

Answer 1

You're right to be worried about this point in the proof. In fact, what you're trying to prove is false: you have the wrong definition of the closure operator!

In a strongly minimal set $D$ defined by a formula with parameters $\varphi(x,\overline{a})$, the closure operator should be $X \mapsto \text{acl}_{\overline{a}}(X) = \text{acl}(X\cup\overline{a})\cap D$.

Here's a counterexample to exchange for your definition of closure, showing that including the parameters is necessary:

Let $L = \{E,f\}$, where $E$ is a binary relation and $f$ is a unary function. Let $T$ be the theory asserting that $E$ is an equivalence relation with infinitely many infinite classes, and $f$ chooses a representative for each $E$-class. That is, $\forall x\, (x\, E\, f(x))$, and $\forall x\, \forall y\, (x\, E\, y) \rightarrow (f(x) = f(y))$.

$T$ is a complete theory with quantifier-elimination. Let $\mathbb{C}$ be its monster model. For any $a\in \mathbb{C}$, the $E$-class of $a$, defined by $x\, E\, a$, is strongly minimal. Call it $D_a$. Note that $\text{acl}(\emptyset) = \emptyset$, since every element of $\mathbb{C}$ has infinitely many conjugates under automorphisms of $\mathbb{C}$ (permuting the equivalence classes).

Choose any $b\in D_a$ such that $f(b)\neq b$, and let $c = f(b)$. Now $c\in (\text{acl}(b)\cap D_a)\setminus (\text{acl}(\emptyset)\cap D_a)$, but $b\notin \text{acl}(c)\cap D_a$, since $b$ has infinitely many conjugates under automorphisms of $\mathbb{C}$ fixing $c$ (take any permutation of $D_a\setminus \{c\}$).

What went wrong, of course, is that $c$ should really be in the closure of $\emptyset$. Indeed, following the correct definition of closure, $c\in \text{acl}(\emptyset \cup \{a\}) \cap D_a$. It's the unique element of $\mathbb{C}$ satisfying the formula $(x\, E\, a) \land (f(x) = x)$.

It's easier to think about / work with strongly minimal sets defined over $\emptyset$, so we often add the parameters $\overline{a}$ to the language at the start of the discussion. Note that if there are constants for $\overline{a}$ in the language, then $\text{acl}(X\cup\overline{a})\cap D = \text{acl}(X)\cap D$.

The popular textbooks by Marker and Hodges are both unfortunately sloppy on this point. Hodges only proves exchange over base sets $A$ containing the parameters $\overline{a}$, but he doesn't explicitly write down the definition of closure in strongly minimal sets defined with parameters. The book by Tent & Ziegler, as usual, gets it right.