# Least model over a certain configuration

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 $\bigoplus_{i < \omega} \mathbb Z/4\mathbb Z$).

• Dont we have the following by uncountably categorical: you have some $M_0$-definable strongly minimal set $\phi(x)$. Now $M_i$ is minimal over $M_0\cup \phi(M_i)$ and $\phi(acl(M_1\cup M_2))$ is strongly minimal. – TimZ May 19 '17 at 11:12
• That is true. Does it help? – Levon Haykazyan May 19 '17 at 12:06
• Now I think that we can show that there is some model N with $\phi(acl(M_1\cup M_2))=\phi(N)$. As $N$ is minimal over $M_0\cup \phi(N)$ we have then that $N= acl(M_1\cup M_2)$. – TimZ May 19 '17 at 12:23
• I think if we take $N$ to be minimal over $M_1 \cup M_2$, then $\phi(acl(M_1M_2)) = \phi(N)$. However $N$ is not algebraic over $\phi(N)$ (unless the theory is almost strongly minimal). I don't see how to get $N = acl(M_1M_2)$. – Levon Haykazyan May 19 '17 at 12:54
• You are right this is not obvious, I thought that $acl(M_1M_2)$ would be would be maximal constructible over $\phi(acl(M_1M_2))$. I dont have any proof. – TimZ May 19 '17 at 14:43

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 $M_1M_2$.
• Nice example. Doesn't this construction work for any group $G$ (instead of $V$)? – TimZ May 22 '17 at 9:02