I'm reading David Marker's Model Theory: an introduction, and in particular theorem 6.1.11:
Theorem 6.1.11: For a complete, countable first order theory $T$, suppose $\mathcal{M}, \mathcal{N}\vDash T$, $\psi(x)$ a strongly minimal $\mathcal{L}(A)$-formula where $A \subseteq M_0$ where $M_0$ is a universe of an elementary substructure of both $\mathcal{M}$ and $\mathcal{N}$. If $\psi(\mathcal{M})$ and $\psi(\mathcal{N})$ have the same dimension, then there is a bijective partial elementary map $f:\psi(\mathcal{M}) \to \psi(\mathcal{N})$.
The proof requires a lemma:
Lemma 6.1.6: Suppose $\mathcal{M}, \mathcal{N}, T, \psi$ and $A$ as above. If $a_1, \cdots, a_n \in \psi(\mathcal{M})$ are independent over $A$ and $b_1, \cdots, b_n \in \psi(\mathcal{N})$ also independent over $A$, then $$tp^\mathcal{M}((a_1, \cdots, a_n)/A)=tp^\mathcal{N}((b_1, \cdots, b_n)/A)$$
The proof of the lemma is easy to follow.
The proof of theorem 6.1.11 begins with this:
Let $B, C$ be bases of $\psi(\mathcal{M}), \psi(\mathcal{N})$ respectively. $B, C$ are assumed to have the same cardinality, hence let $f':B \to C$ be a bijection. By lemma 6.1.6, $f'$ must be elementary.
What I don't understand is why does lemma 6.1.6 apply here necessarily. $B, C$ are independent sets, though not necessarily over $A$, question: could someone clarify the claim that $f'$ must be elementary?
Definitions:
- We assume that $(\psi(\mathcal{M}), Cl)$ defines a pregeometry where $Cl$ is the closure operator defined by $Cl(A)=acl(A) \cap \psi(\mathcal{M})$. $acl$ is the algebraic closure operator where $acl(A)=\{a \in M: a \ \mbox{satisfies an algebraic} \ \mathcal{L}(A)\mbox{-formula of one free variable}\}$. Algebraic formulas are just formulas which define a finite set in $\mathcal{M}$ (and also $\mathcal{N}$ - any definable finite sets are finite in both structures by elementary equivalence).
- A set $X \subseteq \psi(\mathcal{M})$ is independent over $Y \subseteq \psi(\mathcal{M})$ if for any $x \in X, x \notin Cl(Y \cup (X \setminus \{x\}))$. If $Y$ is the empty set then we say $X$ is independent.