I am trying to understand and fill the gaps in the following proof that, over algebraically closed fields of characteristic 0, the model theoretic algebraic closure is equivalent to the field-theoretic algebraic closure.
Let $\mathcal{A}\vDash T_{acf0}$, then by the completeness of $T_{acf0}$ we know there is a $\kappa$-saturated, $\kappa$-homogeneous model $\mathcal{B}$ such that $\mathcal{A}\preceq \mathcal{B} $ and $\kappa>|\mathcal{A}|$. I denote the model theoretic closure by $acl(A)=\{x\in \mathcal{A} : x \text{ is algebraic over } A \} $ and the field-theoretic closure by $acl_f(A)=\{x\in \mathcal{A} : P(x)=0 \text{ for } 0\neq P\in \langle A\rangle[x]\}$. As both these sets are preserved under elementary extensions, it suffices to prove that they coincide in $\mathcal{B}$.
The proof I am studying then proceeds as follows:
(i) First, for all $b,c\in \mathcal{B}\setminus acl_f(A)$ we have that $tp(b/A)=tp(c/A)$, hence $acl(A)\subseteq acl_f(A)$.
(ii) For the converse direction, if $a\in acl_f(A)$, then if $g\in Aut(\mathcal{B}/A)$, then we have $g(a)\in acl_f(a)$ as well. Then the type $tp(a/A)$ has strictly less than $\kappa$ many realizers in $\mathcal{B}$, as $\mathcal{B}$ is $\kappa$-homogeneous. Now, since $\mathcal{B}$ is also $\kappa$-saturated, it follows by compactness that there is some $\phi(x,\overline{b})\in tp(a/A)$ such that $\phi(\mathcal{B},\overline{b})$ is finite.
This is my attempt to fill in the details:
(i) By quantifier elemination, every formula of $T_{acf0}$ is a Boolean combination of atomic formulas. Now, if $b,c\in \mathcal{B}\setminus acl_f(A)$ and $tp(b/A)\neq tp(c/A)$, there is (wlog) an atomic formula $\phi$ such that $\phi \in tp(b/A)$ and $\neg \phi\in tp(c/A)$. Now, atomic formulas in the language of fields can be represented as polynomials, hence, for some $P\in\langle A \rangle[x]$, we either have that $P(b)=0$ or that $P(a)=0$, contradicting $b,c\in \mathcal{B}\setminus acl_f(A)$. Hence $tp(b/A)=tp(c/A)$. Then, for any $b\notin acl_f(A)$ and any $\phi\in tp(b/A)$, there are $|\mathcal{B}\setminus acl_f(A)|=\kappa $ many realizers of $\phi$, hence $b\notin acl(A)$.
(ii) Firstly, notice that any partial map $h:\mathcal{B}\to\mathcal{B}$ such that $h(a)=b$ with $b\in tp(a/A)$ is a partial elementary map, hence by $\kappa$-homogeneity $h$ can be extended to a different automorphism $h'\in Aut(\mathcal{B}/A)$. Hence every realizer of $tp(a/A)$ determines a automorphism fixing $A$. Moreover, since $a\in acl_f(A)$ entails $g(a)\in acl_f(a)$ for any $g\in Aut(\mathcal{B}/A)$, it follows that there are at most $|acl_f(A)|$ values for $a$ for automorphisms fixing $A$. By the above observation this then means that there are at most $|acl_f(A)|<\kappa$ realizers of $tp(a/A)$.
Now, and this is the part I am more doubtful about, it follows by $\kappa$-saturation that the set of realizers of $tp(a/A)$ is actually finite. Why? Because otherwise we could apply compactness and show that the type obtained adding to $tp(a/A)$ formulas $x\neq d_i$ for all $d_i\in tp(a/A)$ is consistent, hence by $\kappa$-saturation it has a realizer $q\in \mathcal{B}\setminus tp(a/A)$, contradiction. Therefore, it follows that the set of realizers of $tp(a/A)$ is finite, hence by compactness there is a formula $\phi(x,\overline{b})\in tp(a/A)$ such that $\phi(\mathcal{B},\overline{b})$ is finite.
Is this correct?
Your elaboration of the proof seems exactly right to me.
The suggested proof of (ii) is a rather bizarre way of proving $\text{acl}_f(A)\subseteq \text{acl}(A)$, though. Here is a more elementary argument.
Suppose $b\in \text{acl}_f(A)$. Let $K = \langle A\rangle$ be the subfield generated by $A$. Then $b$ is a root of some non-zero polynomial $p(x)\in K[x]$. The condition $p(x) = 0$ can be written as a formula with parameters from $A$, and it is satisfied by at most $\text{deg}(p)$ elements, so it is an algebraic formula. Thus $b\in \text{acl}(A)$.