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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):

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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.

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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.

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