Does every uncountably categorical theory have a $\varnothing$-definable strongly minimal imaginary?

An uncountably categorical theory always has a strongly minimal set definable over its prime model, but sometimes this set needs parameters to define.

By a $$\varnothing$$-definable imaginary I mean the quotient of some finite power of the home sort by a $$\varnothing$$-definable equivalence relation. No added generality is gained by considering quotients of definable sets, since we can always just extend the equivalence relation and send everything in the complement to a single point.

Every example of an uncountably categorical theory I can think of has a $$\varnothing$$-definable strongly minimal imaginary. So the question is:

Does every uncountably categorical theory have a $$\varnothing$$-definable strongly minimal imaginary?

The best I've been able to come up with is the theory of $$\mathbb{Z}$$ with a symmetric successor relation and a predicate for $$x\equiv y\text{ (mod 2})$$. By a direct argument you can show that no $$\varnothing$$-definable equivalence relation on elements gives a strongly minimal imaginary, but I believe there is a strongly minimal quotient of 2-tuples, specifically unordered adjacent pairs.

A few easier questions that would be relevant would be the same but allowing $$\text{acl}^\text{eq}(\varnothing)$$-definable imaginaries or only requiring that the imaginary have Morley rank $$1$$, but allowing it to have Morley degree greater than 1.

I honestly thought about a lot of different possible counterexamples before realizing that a small modification of the structure I already mentioned gives a counterexample. Specifically if we consider the structure of $$\mathbb{Z}$$ with the successor function and a predicate $$P(x,y)$$ for $$x\equiv y\text{ (mod 2})$$. Let $$T$$ be the theory of this structure.

To see this assume that $$T$$ has a $$\varnothing$$-definable strongly minimal imaginary $$I$$. Since the theory in uncountably categorical there must be a definable finite-to-finite correspondence between $$I$$ and the definable set $$P(x,0)$$ (where $$0$$ is some element of the prime model). Let $$\varphi(x,y,\overline{a})$$ be that finite-to-finite correspondence (and $$\overline{a}$$ is a tuple of parameters in the home sort). There needs to be a uniform finite bound $$n$$ on the sizes of the fibers of the correspondence, so we can write a formula $$\psi(w,\overline{z})$$ that means '$$\varphi(x,y,\overline{z})$$ defines a finite-to-finite correspondence between $$I$$ and $$P(x,w)$$ with fibers of size at most $$n$$'. Then the theory $$T$$ says $$\forall w \exists \overline{z} \psi(w,\overline{z})$$, so in particular there is some set of parameters $$\overline{a}$$ in the prime model such that $$\psi(x,y,\overline{a})$$ is a finite-to-finite correspondence between $$I$$ and $$P(x,0)$$. Since all of the prime model is in $$\text{dcl}(0)$$, we can just let $$\overline{a}$$ actually be $$0$$, so we have a formula $$\psi(x,y,0)$$ that defines a finite-to-finite correspondence between $$I$$ and $$P(x,0)$$.

There needs to be a number $$m$$ such that for any $$b\in I$$, the largest and smallest element of the home sort related to $$b$$ by $$\varphi(x,y,0)$$ are no more than $$m$$ many successor steps away, (EDIT: There's a subtlety at this point which is that the set related can be arbitrarily large (it could always contain 0 for instance), but the point is it needs to generically contain a chunk around 0 of bounded size and some moving chunk of bounded size. You can define the needed function in terms of the moving chunk.) so we can actually define a formula $$\chi(x,y,0)$$ such that each element of $$I$$ is related to the least element it was related to by $$\varphi(x,y,0)$$, or $$0$$ if there were no elements related (this can only happen for finitely many elements of $$I$$), in other words $$\chi(x,y,0)$$ defines a function $$f(0,x)$$ from $$I$$ into the set $$P(x,0)$$. The image of this function must be all but finitely many elements of the set $$P(x,0)$$.

By symmetry, for any even number $$2k$$, $$f(2k,x)$$ must define an almost surjective map from $$I$$ to the set $$P(x,0)$$. Fix some $$2k$$. I claim that for all but finitely many $$b \in I$$, $$f(0,x)=f(2k,x)$$. If this weren't true, then $$T$$ would interpret the group $$(\mathbb{Z},+)$$, which it can't because it's $$\omega$$-stable. So we can define a new function over the imaginary parameter corresponding to the $$P$$-equivalence class of $$0$$, which I'll call $$[0]_P$$. The function, $$g([0]_P,x)$$, is given by the average behavior of the function $$f(2k,x)$$ for different values of $$2k$$. By stability and symmetry we only need to check some sufficiently large collection of parameters, so the definition of $$g([0]_p,x)$$ is something like 'there exists more than 17 elements $$y$$ of $$[0]_P$$ such that $$f(y,x)$$ outputs the same value for, let $$g([0]_P,x)$$ be that value'.

By symmetry again everything we said about the even numbers works for the odd numbers as well and we have a definable function $$g([1]_P,x)$$ from $$I$$ into the set $$P(x,1)$$. Since $$[1]_P$$ is obtained from $$[0]_P$$ by shifting everything up by $$1$$, it must be the case that $$g([1]_P,x)=S(g([0]_P,x))$$, but by symmetry we also have $$g([0]_P,x)=S(g([1]_P,x))$$, which is impossible. Therefore no such imaginary can exist.

There's still the matter of the followup question regarding $$\varnothing$$-definable imaginaries with finite Morley rank, but that's not the title of this question so I'm posting this as an answer rather than as an edit to the question.

• Very nice example! – Alex Kruckman Feb 15 at 15:21