The question is as in title. By acl-dimension I understand the cardinality of maximal acl-independent set (well-defined for strongly minimal theories). By minimal I understand that there is no equivalent model of smaller dimension.

It is easy to find examples where it is 0, 1 or $\aleph_0$ (algebraic closure of rationals, rationals as a vector space over themselves, and a countably infinite dimensional vector space over a finite field), and any case can be reduced to 0 or $\aleph_0$, possibly by adding finitely many constant symbols (names for the elements of a basis), and by of course it cannot be more than $\aleph_0$, but I can't seem to think of an example with minimal dimension which is finite, but greater than 1.

I believe it can also be shown that in a strongly minimal model, any infinite, algebraically closed subset is the universe of an elementary substructure, so the question can be restated as the following: what are the possible minimal dimensions of infinite sets in strongly minimal models?


2 Answers 2


Here is an example of a strongly minimal theory whose prime model has dimension $2$. Let the language $\cal L$ contain just one ternary function symbol $\oplus$. Consider the $\cal L$-theory of $\mathbb Q$, where $\oplus$ is interpreted as $\oplus(x, y, z) = x + y - z$. We also use the infix notation $x \oplus_z y$. Let's look at this theory. In particular it would say

  • for every $z$ the domain is a non-trivial divisible torsion-free abelian group with identity $z$ under operation $\oplus_z$;
  • $\forall x, y, z, w, u (\oplus(x, y, z) = u \leftrightarrow \oplus(x, y, w) = \oplus(z, u, w))$

Denote the last theory by $T$. Now $T$ is uncountably categorical. Indeed given any two uncountable models ${\cal M}_1$ and ${\cal M}_2$ of the same cardinality, pick any two elements $w_1 \in M_1$ and $w_2 \in M_2$. Then $M_i$ is a $\mathbb Q$ vector space under $\oplus_{w_i}$. So there is an isomorphism $\alpha$ of structures $\langle M_i, \oplus_{w_1}, w_1\rangle$ and $\langle M_i, \oplus_{w_2}, w_2\rangle$. But then this isomorphism preserves $\oplus$ as the last axiom holds in both structures. In particular $T$ is complete and strongly minimal as an uncountable model of $T$ (which is $\aleph_0$-saturated) is definably interpretable in a strongly minimal set (a $\mathbb Q$ vector space).

Now the prime model of $T$ is $\mathbb Q$. But since both $x \mapsto \lambda \cdot x$ and $x \mapsto x + \lambda$ are automorphisms, we can construct an automorphism that maps any given pair $(x_1, y_1)$ with $x_1 \neq y_1$ to any given pair $(x_2, y_2)$ with $x_2 \neq y_2$. So for every $q \in \mathbb Q$ we have $acl(q) = \{q\}$. Also since $acl(0, 1) = \mathbb Q$ the prime model has dimension $2$.

This is a correction of an exercise in Marker's book by Richard Rast. I think he has taken down the original, so I can't put a link to it.


In Example 5.2 in Hrushovski's original paper, the prime model has dimension 3. I add; the reason is the algebraic closure of a set of size 2 has cardinality at most 3.

  • 1
    $\begingroup$ Do you mean the section 5.2 of "A new strongly minimal set" (the paper disproving Zilber's conjecture)? If not, which paper do you mean? $\endgroup$
    – tomasz
    Commented Jun 28, 2018 at 23:16
  • $\begingroup$ Yes, that is the example I mean. $\endgroup$ Commented Jul 28, 2021 at 2:21

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