Least model over a certain configuration

Question

I am interested in conditions under which there is a least model inside a (not saturated in general) model $N$ over certain configurations such as $M_0 \subseteq M_1, M_2$ with $M_1 \overset{\vert}{\smile}_{M_0} M_2$. The situation is simpler in an uncountably categorical theory, since $N$ has to be saturated and the intersection of all models containing $M_1M_2$ is $acl(M_1M_2)$.

Question: given $M_0 \subseteq M_1, M_2$ models with $M_1 \overset{\vert}{\smile}_{M_0} M_2$ inside the monster model of an uncountably categorical theory, is $acl(M_1M_2)$ a model?

On the face of it I don't see many reasons why it should be true. However, it is certainly true in almost strongly minimal theories and also in the classical example of an uncountably categorical theory that is not almost strongly minimal (the theory of $\bigoplus_{i < \omega} \mathbb Z/4\mathbb Z$).

Answer 1

I think I have found the counterexample I was looking for.

The idea is to hack the example $\bigoplus_{i < \omega} \mathbb Z/4 \mathbb Z$ to keep it not almost strongly minimal, but to remove the feature that makes $acl(M_1M_2)$ into a model. What is happening is there is a formula $\phi(x) = 2x = 0$ which defines a vector space over $\mathbb Z/2\mathbb Z$ and hence is strongly minimal. Above each $b \in \phi(M)$ there is a set defined by $\psi(x, b) = 2x = b$ which is internal to $\phi$ but every element realises the same type so is not algebraic. When we have two models $M_1$ and $M_2$, $acl(M_1M_2)$ adds new elements to $\phi(M_1) \cup \phi(M_2)$, e.g. $m_1 + m_2$ for $m_i \in M_i$. So there need to be new elements in $\psi(x, m_1+m_2)$. However such an element can be found by taking $n_1+n_2$ where $n_i$ is in $\psi(x,m_i)$ which is therefore in $acl(M_1M_2)$. The idea is to remove the interaction between $\psi(x,b)$ for different $b$-s and so to remove the possibility of finding such an element in $acl(M_1M_2)$.

So the counterexample is as follows: There are two sorts $R$ and $V$. The sort $V$ is a vector space over $\mathbb Z/2\mathbb Z$ so is strongly minimal. $R = V^2$ but the group operation on $V^2$ is not in the language. Instead there is a projection map $\pi \colon R \to V, (a,b) \mapsto b$ onto the second coordinate and also a map $\sigma : R \times V \to R, ((a,b), c) \mapsto (a+c, b)$ describing the action of $V$ on each fibre $\pi^{-1}(b)$. Now fibres of $\pi$ are internal to $V$ as in the above example. However if we now pick two models $m_1 \in M_1$ and $m_2 \in M_2$, then the type of an element in $\pi^{-1}(m_1+m_2)$ is not algebraic over $M_1M_2$.