Definable in a theory $T$ strongly minimal

Question

Let T strongly minimal and $\phi(x,y)$ a $A$-formula and $b$ an arbitrary element. Let $\mathbb{M}$ the monster model. Suppose $\phi(\mathbb{M},b)$ is quasi-definable over $A$. Show that $\phi(\mathbb{M},b)$ is definable over $acl(A \cup \{e\})$ for all $e \notin acl(A)$.

A set X is quasi-definable over $A$ if is definable over all model $M$ containing $A$.

The idea is find $\alpha $ a $L(A \cup \{e\})$-formula that define $\phi(\mathbb{M},b)$ considering that $e \notin acl(A)$ then for all $\beta $ $L(A)$-formula such that $\mathbb{M}\models \beta (e)$ then $\mid \beta(\mathbb{M})\mid \ge \aleph_0 $ so $\alpha$ this formula should depend of $\beta$ and it's infinites realizations in the monster model, but not how to build it and and where it is needed that T is strongly minimal. Thanks for any hint.

Answer 1

As none has answered so yet, I post my answer which I feel is too long and involved.

Assume $T$ is complete theory with an infinite model. Then the following are equivalent (this I assume without proof)

1. $\phi(\mathbb{M},b)$ is quasi-definable over $A$;
2. the orbit of $\phi(\mathbb{M},b)$ over $A$ is finite;
3. there is a definable equivalence relation $\varepsilon(y,z)\in L(A)$ that partition $\mathbb{M}$ into finitely many classes and $\phi(\mathbb{M},b)=\phi(\mathbb{M},c)$ whenever $\varepsilon(c,b)$ and $c\equiv_Ab$.

Now back to your question.

Assume first that $\varepsilon(\mathbb{M},b)$ is finite. Then it is contained in every model containing $A$ hence it is a subset of ${\rm acl}(A)$ and the claim is trivial.

Now assume $\varepsilon(\mathbb{M},b)$ is infinite and let $T$ be strongly minimal. Then $\varepsilon(\mathbb{M},b)$ is co-finite. As any $e\notin{\rm acl}(A)$ is conjugate to $b$, also $\varepsilon(\mathbb{M},e)$ is co-finite as-well. Then $\varepsilon(e,b)$ so, by (3) above, we obtan $\phi(\mathbb{M},b)=\phi(\mathbb{M},e)$.