# Marker Lemma 6.1.20 - Categoricity Theorem

This is used as a lemma to prove a weaker version of the Baldwin-Lachlan Theorem (assuming the existence of a strongly minimal L-formula):

What's unclear to me is the line where Marker says "It is easy to see that...". I cannot figure out what property of omega saturation or what lemma he invoked to arrive at this step.

## 1 Answer

Suppose for a contradiction that $$\dim(\phi(\mathcal{M}^*)) = k < \aleph_0$$. Then there are independent elements $$a_1, \ldots, a_k$$ of $$\phi(\mathcal{M}^*)$$, such that $$\phi(\mathcal{M}^*) \subseteq \operatorname{acl}(a_1, \ldots, a_k)$$. Now consider: $$\Sigma(x) = \{\phi(x)\} \cup \{\neg \psi(x) : \psi(x) \in \mathcal{L}(a_1, \ldots, a_k) \text{ an algebraic formula}\}.$$ Then $$\Sigma(x)$$ is finitely satisfiable, because any finite subset of $$\Sigma(x)$$ will only exclude finitely many elements from $$\phi(\mathcal{M}^*)$$. So because $$\mathcal{M}^*$$ is $$\omega$$-saturated, there must be some realisation $$a \in \mathcal{M}^*$$ of $$\Sigma(x)$$. By construction of $$\Sigma(x)$$ we then have that $$a$$ is not algebraic over $$a_1, \ldots, a_k$$, but we still have $$a \in \phi(\mathcal{M}^*)$$. This contradicts $$\phi(\mathcal{M}^*) \subseteq \operatorname{acl}(a_1, \ldots, a_k)$$, so we conclude that $$\dim(\phi(\mathcal{M}^*))$$ cannot be finite.