Geometric structures and prime models I found the following two definitions in a survey by D. Macpherson. 
A structure $\mathcal{M}$ is uniformly bounded if, given a definable family of subsets of $M$, there is an integer $N$ such that whenever an element of the family is a finite set, its cardinality is bounded from above by $N$. 
A structure is geometric if it is uniformly bounded and if $acl$ defines a pregeometry in every model of $Th(\mathcal{M})$.
Clearly strongly minimal and o-minimal structures are geometric in this sense.
Fix a geometric structure $\mathcal{M}$ (in a countable language, if necessary). My question: for $A\subseteq M$, does $acl(A)$ define a (unique?) prime model over $A$ of $Th(\mathcal{M})$? 
The proof I've seen for the o-minimal case is ad hoc. 
Maybe the right question to ask is: are there always (unique) prime models over $A$, if $M$ is geometric (the answer is yes for strongly minimal models by $\omega$-stability)?
More generally, I was surprised not to find the notion of geometric structure in basic model theory books. It seems to capture what strongly minimal and o-minimal theories have in common, in particular the notion of dimension coming from the pregeometry $acl$. Now, I realise that while in the strongly minimal case the theory of dimension leads to a complete classification of the models, in the o-minimal case the situation is much more complicated. However, I was wondering how far one could push the analogy between these two types of theories, based only on the uniform finiteness of definable families and the fact that $acl$ is a pregeometry. Hence, my question about prime models.
 A: This is not even true for strongly minimal structures in general. For example, let $T$ be the theory of infinite-dimensional $\mathbb{F}_p$-vector spaces. Then for any finite $A$, $\text{acl}(A)$ is the span of $A$, which is finite dimensional, and hence not a model of $T$. 
In general: Every strongly minimal theory $T$ has a prime model (over $\emptyset$) of dimension $d \leq \aleph_0$, and it is true that if a set $A$ contains an algebraically independent set of size $d$, then $\text{acl}(A)$ is a prime model over $A$. But if $\dim(A)<d$, $\text{acl}(A)$ fails to be a model of $T$. In the example above, $d = \aleph_0$.
For a reference, see the Claim in the proof of Theorem 5.7.8 on p.85 of Tent & Ziegler. This claim says that any infinite algebraically closed subset of a model $M$ is an elementary substructure. From this, it follows that for any $A$ such that $\text{acl}(A)$ is infinite, $\text{acl}(A)$ is a prime model over $A$ (since for any theory, any partial elementary map $f\colon A\to M$ extends to a partial elementary map $\text{acl}(A)\to M$).  So it remains to see that the condition "$\text{acl}(A)$ is infinite" agrees with the condition I stated above, "$\dim(A)\geq d$, where $d$ is the dimension of the prime model". This is because $\text{acl}(A)$ is infinite iff $\text{acl}(A)$ is a model (all models of $T$ are infinite) iff the prime model elementarily embeds in $\text{acl}(A)$ iff $\dim(A) = \dim(\text{acl}(A)) \geq d$.
You also bring up the fact that o-minimal theories are only guaranteed to have definable Skolem functions when they expand ordered groups. This gives another counterexample in the o-minimal setting: In the theory of dense linear orders without endpoints, $\text{acl}(A) = A$ for all sets $A$, so $\text{acl}(A)$ is not a model unless $A$ is already. 
Finally, it is not true that every geometric theory admits prime models over sets - the geometric hypothesis is too weak. I've now remembered a rather silly construction: Given any structure $M$ in a relational language, add a new binary relation symbol $E$ to the language $L$ and create a new structure $M'$ by replacing every element of $M$ by an infinite $E$-equivalence class. Then we have a quotient map $q\colon M'\to M$, and we set $M'\models R(a_1,\dots,a_n)$ if and only if $M\models R(q(a_1),\dots,q(a_n))$ for every $n$-ary relation symbol $R$ in $L$.
The "blow-up" structure $M'$ has trivial algebraic closure ($\text{acl}(A) = A$ for all $A$), so it's trivially geometric. But if $\text{Th}(M)$ doesn't have prime models over sets, neither does $\text{Th}(M')$.
Of course, we could add the hypothesis of definable Skolem functions, but this is overkill, since then we very easily get that $\text{acl}(A)$ is prime over $A$ for any set $A$ without using the geometric hypothesis.
