# 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.

• "I was surprised not to find the notion of geometric structure in basic model theory books." This doesn't seem surprising to me - the basic model theory books I'm familiar with don't even include the notion of o-minimal structure! – Alex Kruckman Dec 14 '17 at 19:05
• "The proof I've seen for the o-minimal case is ad hoc." I'm curious what proof you have in mind. The proof I know goes by showing that o-minimal theories admit definable Skolem functions. Having definable Skolem functions is the easiest way to get prime models over sets, and seems to me not at all ad hoc. – Alex Kruckman Dec 14 '17 at 19:06
• @AlexKruckman “the basic model theory books ... don't even include the notion of o-minimal structure!”... Marker, Hodges, Rothmaler, Marcija-Toffalori do. True, the more recent Tent-Ziegler doesn't and of course no book written before the '80s does! Anyway, what I meant is that it would seem fairly natural to me to introduce strongly minimal and o-minimal theories side by side, but maybe, at least at textbook level, such an analogy does not go very far, so the notion of geometric structure is not that relevant. This was the sense of my question about prime models. – R. Harlow Dec 15 '17 at 10:39
• Regarding the “ad hoc” argument, my remark was confused. There are two things: the existence and uniqueness of prime models over sets for all strongly minimal and all o-minimal theories (which, in the o-minimal case, does use intervals and endpoints) and the possibility to realise such a prime model as the acl of something (which uses the existence of definable Skolem functions). Now, in the o-minimal case, the existence of definable Skolem functions is guaranteed only if the structure expands an ordered group. I don't know if every strongly minimal structure has definable Skolem functions. – R. Harlow Dec 15 '17 at 10:42
• "I don't know if every strongly minimal structure has definable Skolem functions". Consider an equivalence relation with infinitely many classes, all of size $3$. – Alex Kruckman Dec 15 '17 at 13:51

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.

• I haven't worked out yet why your “In general:...” paragraph is true (do you have a reference or a hint to give me?). I realise now that I don't know if acl(A) is a prime model of an o-minimal theory which does not expand an ordered group...so maybe even the rescued version of my original question on acl (the one you suggest at the end of your answer) is too much to ask without adding further hypotheses (i.e. the existence of definable Skolem functions). – R. Harlow Dec 15 '17 at 10:49
• @R.Harlow I've just made significant edits to my answer. – Alex Kruckman Dec 15 '17 at 14:26
• This settles the matter completely. Thank you very much. – R. Harlow Dec 17 '17 at 8:03