Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

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.

share|improve this answer

If you consider an elementary extension saturated enough, you can find sequences $\{a_i\}$ of independent elements of any cardinality: just keep finding realisations of the unique non-algebraic 1-type over $\{a_i\}_{i < \alpha}$ (incidentally, this has a name, such sequnces are called Morley sequences).

So the question really boils down to finding out how saturated is the particular model you are looking at.

share|improve this answer
That's not what the question is about. What you suggest would supply models of arbitrarily large dimension, but that is easy enough, as you say (and you don't even need to use Morley sequences for that, upwards Skolem is enough). I'm asking about strongly minimal models (as in, models whose Morley rank is $1$) of arbitrary (or even any other than $0,1$) finite dimension with no elementary submodels, if they exist at all. –  tomasz Aug 7 '12 at 0:29
(Upwards Lowenheim-Skolem is not enough; it just gets you big models, but doesn't tell you how many independent elements are there) –  Dima Sustretov Aug 7 '12 at 0:46
It is enough. For infinite $\kappa$, algebraic closure of a set of cardinality $\kappa$ (in countable theory) has cardinality $\kappa$, so the basis of a model of uncountable cardinality $\kappa$ is necessarily of cardinality $\kappa$ itself. –  tomasz Aug 7 '12 at 0:49
ah, ok, so you suppose the theory is countable. –  Dima Sustretov Aug 7 '12 at 0:56
I still maintain that the question is really about how saturated the given (minimal) model is. Which seems (at least to me) to be hard to find out without knowing something about the particular theory. –  Dima Sustretov Aug 7 '12 at 1:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.