$\omega$- categoricity and algebraic closure

Question

1. Let $\mathcal{M}$ be an infinite structure whcih is also $\omega$-categorical. Let a finite subset $A\subset M$ be given. Under what condition we can conclude that $A\subseteq acl^{\mathcal{M}}(\emptyset)$?

2. What does $\omega$-categoricity tell us about $acl^{\mathcal{M}}(\emptyset)$, in general?

Answer 1

Question 2: It tells us that $\text{acl}^{\mathcal{M}}(\emptyset)$ is finite. More generally, if $\mathcal{M}$ is any model of an $\omega$-categorical theory and $A\subseteq \mathcal{M}$ is finite, then $\text{acl}^M(A)$ is finite.

Proof: Suppose $A = \{a_1,\dots,a_n\}$. If $b\in \text{acl}(A)$, then this is witnessed by an algebraic formula $\varphi(x,a_1,\dots,a_n)$. By Ryll-Nardzewski, there are only finitely many formulas in the variables $x,y_1,\dots,y_n$ up to equivalence, and each algebraic formula has only finitely many solutions, so $\text{acl}(A)$ is finite.

Question 1: As Noah says in this comments, it's not clear what kind of answer you're hoping for here. But here is one way to answer the question: If $\mathcal{M}$ is the unique countably infinite model of an $\omega$-categorical theory (or more generally if it is a strongly $\omega$-homogeneous model of this theory), then $A\subseteq \text{acl}^\mathcal{M}(\emptyset)$ if and only if $A$ is finite and the orbit of $A$ under the action of $\text{Aut}(\mathcal{M})$ is finite.

Proof: Suppose $A\subseteq \text{acl}(\emptyset)$. Since $\text{acl}(\emptyset)$ is finite, $A$ is finite. For each $a\in A$, there is some algebraic formula $\varphi(x)$ satisfied by $a$, and hence by every element in the orbit of $a$ under the action of $\text{Aut}(\mathcal{M})$. So the orbit of $a$ is finite. Since $A$ has finitely many elements, each of which has finite orbit, the orbit of $A$ as a set is also finite.

Conversely, suppose $A\not\subseteq \text{acl}(\emptyset)$. Then there is some $a\in A$ with $a\notin \text{acl}(\emptyset)$, so $\text{tp}(a/\emptyset)$ has infinitely many realizations. And for any $a'$ realizing $\text{tp}(a/\emptyset)$, there is an automorphism $\sigma\in \text{Aut}(\mathcal{M})$ with $\sigma(a) = a'$, since $\mathcal{M}$ is strongly $\omega$-homogeneous. It follows then that $A$ is infinite or has infinite orbit as a set.