Testcases for Baldwin Lachlan

Question

Let $T$ be a complete theory in a countable language with infinite models. By a theorem of Baldwin-Lachlan, $T$ is uncountably categorical if and only if it is $\omega$-stable and has no Vaughtian pairs.

The proof works roughly as follows:

• Show that there is a prime model $\mathcal{M}_0$ of $T$
• Prove that there is a strognly minimal formula $\phi$ with parameters in $\mathcal{M}_0$
• Prove that the isomorphism type of a model $\mathcal{N}$ is determined completely by the dimension of $\phi^{\mathcal{N}}$ over $\mathcal{M}_0$

The prototypical examples are of algebraically closed fields of a fixed characteristic and vector spaces over a countably infinite field. Here there are the familiar notions of dimension in the form of transcendence degree or the usual dimension from linear algebra. In these two cases the strongly minimal formula is just $x=x$, so the way the strongly minimal set sits inside the model is not very interesting (it is everything!).

A slightly more interesting example is given by the theory of the direct sum of infinitely many cyclic groups of order $4$. If $D$ is the set of all elements of order $2$, then $D$ is strongly minimal. The model is not algebraic over $D$, in contrast with the previous two examples.

I am interested in theories that illustrate the difficulty of the theorem, in the spirit of the last example. For instance:

• Is it possible that there is no strongly minimal formula which is definable without parameters?
• Must the strongly minimal set be unique up to perhaps finitely many elements?

Answer 1

The following example answers both questions. Consider a theory $T$ in the language $L = \langle E, f \rangle$, where $E$ is a binary relation, $f$ is an unary functional symbol. The theory $T$ asserts

• $E$ is an equivalence relation with $2$ equivalence classes.
• $f$ pairs up elements from different equivalence classes. That is $f$ is a bijection, $f(f(x)) = x \land \lnot E(x, f(x))$ for all $x$.
• there are infinitely many elements.

It is easy to see that $T$ is uncountably categorical. Each equivalence class is a strongly minimal set (hence there are two strongly many sets). Yet to get your hands on one you need a parameter. Indeed let $\phi(x)$ be a formula defining an infinite set. Pick any element $a$, then $\phi(x) \land E(x, a)$ and $\phi(x) \land \lnot E(x, a)$ partition the set defined by $\phi(x)$. Now observe that $f$ is an automorphism of the structure and hence establishes a bijection between the sets defined by $\phi(x) \land E(x, a)$ and $\phi(x) \land \lnot E(x, a)$. Hence they are both infinite and $\phi$ is not strongly minimal.