Clarification with proof: there is a bijective partial elementary map between two strongly minimal sets of same dimension.

Question

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.

Answer 1

The answer is a silly one: When Marker takes $B$ and $C$ to be bases for $\varphi(\mathcal{M})$ and $\varphi(\mathcal{N})$, respectively, he really means that $B$ and $C$ should be independent over $A$, and then of course the lemma applies.

But the issue of parameters for strongly minimal sets is subtle, and I think Marker sweeps it under the rug a bit too aggressively. When $D$ is a strongly minimal subset of $\mathcal{M}$, defined by a formula $\varphi(x)$ with parameters from $A$, we can pretend that $D$ is definable without parameters by adding the parameters $A$ to the language as constant symbols. Note that doing this doesn't change the fact that $D$ is strongly minimal. But it does change the algebraic closure operator on $D$: now if $B\subseteq D$, an algebraic formula with parameters from $B$ can also use the constant symbols naming $A$ as parameters, so $$\text{acl}_{L(A)}(B) = \text{acl}_L(B\cup A).$$

It turns out that when dealing with strongly minimal sets definable with parameters (which are unavoidable in the proof of the Baldwin-Lachlan theorem), the correct closure operator to consider is $\text{acl}_{L(A)}$. Note that independence with respect to $\text{acl}_{L(A)}$ is exactly independence over $A$ with respect to $\text{acl}_{L}$.

This might worry you, since definable sets don't come with unique parameter sets: if $D$ is definable with parameters from $A$, it's also definable with parameters from $A'$ for any $A\subseteq A'$, and $\text{acl}_{L(A)}$ and $\text{acl}_{L(A')}$ might be different in general - even worse, $D$ might have different dimensions with respect to these two closure operators. But this is ok, as long as we're clear about what set of parameters we're considering whenever we talk about a strongly minimal set: the parameters should always be part of the data. Marker seems to avoid making this explicit, which I view as a flaw in his exposition.