Does algebraicity implies dividing (model theory) Suppose $A$ is a set of parameters in a model $M$ and $a,\overline{b}$ are respectively an element and a tuple from $M$ such that $a\in \operatorname{acl}(A\overline{b})\setminus \operatorname{acl}(A)$. 
It is clear that in this case, the type $\operatorname{tp}(a/A\overline{b})$ forks over $A$: If $\{a_1=a,a_2\ldots,a_k\}$ are the finitely many realizations of the type $\operatorname{tp}(a/A\overline{b})$, then we have $$\operatorname{tp}(a/A\overline{b})\models \bigvee_{i=1}^k x=a_i,$$and each of the formulas $x=a_i$ 2-divide over $A$.
However, my question is: 

Does $\operatorname{tp}(a/A\overline{b})$ necessarily divides over $A$?

 A: The type $\operatorname{tp}(a/Ab)$ necessarily divides over $A$. Let $\phi(x, b)$ define (over $A$) the least finite set $\{a = a_1, ..., a_n\}$ containing $a$. The claim is $\phi(x, b)$ $2$-divides over $A$. 
The argument I found is somewhat complicated and uses Neumann's Lemma: Let $X$ be an infinite set, $G \le \operatorname{Sym}(X)$ and $B \subset X$ be finite. Assume that the orbit of each of $c_1, ..., c_n \in X$ is infinite under $G$, then there is $g \in G$ such that $g \cdot c_i \not \in B$ for all $i = 1, ..., n$. See Exercise 6.1.15 in Tent, Ziegler "A course in model theory".
By this lemma there is $f_1 \in \operatorname{Aut}(\mathfrak C/A) = G$ such that $\{f_1(a_1), ..., f_1(a_n)\} \cap \{a_1, ..., a_n\} = \emptyset$. This of course implies that $\phi(x, f_1(b)) \land \phi(x, b)$ is inconsistent. So we take $b_1 = f_1(b)$. Now applying the lemma again we find $f_2 \in \operatorname{Aut}(\mathfrak C/A)$ such that $\{f_2(a_1), ..., f_2(a_n)\} \cap \{f_1(a_1), ..., f_1(a_n), a_1, ..., a_n\} = \emptyset$ and take $b_2 = f_2(b)$. Continuing this way we obtain a sequence $b = b_0, b_1, b_2, ...$ witnessing the $2$-dividing of $\phi(x, b)$.
I am curious if there is a simpler argument.
A: $\newcommand{\tp}{\operatorname{tp}}\newcommand{\Aut}{\operatorname{Aut}}\newcommand{\acl}{\operatorname{acl}}$Here is an argument without involving (directly?) Neumann's Lemma. It is essentially a solution of Exercise 5.6.2 of Tent-Ziegler's book, and was referenced in the Exercise 6.1.15 mentioned in Levon's answer.
First we start with the following lemma:

Lemma: If $c_1,\ldots,c_n$ are elements such that $\tp(c_i/A)$ is not algebraic, then for any finite set $B$ there are is a tuple $(c_1',\ldots,c_n')$ realising $\tp(c_1,\ldots,c_n/A)$ and disjoint from $B$.

Proof of the Lemma: By induction on $n$.
For $n=1$, since the type $\tp(c_1/A)$ is not algebraic, it has infinitely many realisations, and so there must exists $c_1'\equiv_A c_1$ not in $B$.
Consider now a tuple $(c_1,\ldots,c_n,c_{n+1})$ such that $\tp(c_i/A)$ is not algebraic for each $i\leq n+1$. We have two cases:


*

*$\tp(c_i/Ac_{n+1})$ is not algebraic for every $i\leq n$:
Since $\tp(c_{n+1}/A)$ is not algebraic, we have by the case $n=1$ that there exists $c_{n+1}'\equiv_A c_{n+1}$ such that $c_{n+1}'\not\in B$. Let $\alpha$ be an automorphism over $A$ such that $\alpha(c_{n+1})=c_{n+1}'$. Then, the type $\tp(\alpha(c_i)/Ac_{n+1}')$ is not algebraic for every $i\leq n$, and we have by induction hypothesis that there is a tuple $c_1',\ldots,c_n'\equiv_{Ac_{n+1}'} \alpha(c_1),\ldots,\alpha(c_n)$, which is disjoint from $B$. 
Thus, $(c_1,\ldots,c_n,c_{n+1})\equiv_A (c_1',\ldots,c_n',c_{n+1}')$ and 
$\{c_1',\ldots,c_n',c_{n+1}'\}\cap B=\emptyset$. 


*$\tp(c_i/Ac_{n+1})$ is algebraic for some $i\leq n$:
First, notice that by the induction hypothesis there is a tuple $\overline{c}'=(c_1',\ldots,c_n')\equiv_A (c_1,\ldots,c_n)=\overline{c}$ which is disjoint from $B$. Moreover, applying repeatedly the induction hypothesis, we can show that there are infinitely many tuples $\overline{c}^m\equiv_A \overline{c}$ such that all of the tuples are pairwise disjoint, and all of them are disjoint from $B$. (This can be done by applying the lemma for the sets $B,B\overline{c}^1,B\overline{c}^1\overline{c}^2,\ldots$)
For every $m<\omega$, let $\alpha_m\in \Aut(\mathfrak{C} / A)$ be an automorphism sending the tuple $\overline{c}$ to $\overline{c}^m$. If the conclusion of the lemma does not hold, then in particular $\alpha_m(c_{n+1})\in B$ for all $m<\omega$, and since $B$ is finite, there is $b\in B$ such that $\alpha_m(c_{n+1})=b$ for every $m$ in an infinite subset $J\subseteq \omega$.
Also, since $\tp(c_i/Ac_{n+1})$ is algebraic, it has at most $k$ realizations for some $k<\omega$, and we have  $\tp(\alpha_m(c_i)/A\alpha_m(c_{n+1}))=\tp(\alpha_m(c_i)/Ab)$ has $k$ realizations from every $m\in J$. This is a contradiction because, by construction, $\alpha_m(c_i)\neq \alpha_{m'}(c_i)$ for every $m\neq m'$ in $J$. $\Box$
Now we can prove the desired result:

Proposition: If $a\in \acl(A\overline{b})\setminus \acl(A)$ then $\tp(a/A\overline{b})$ 2-divides over $A$.

Proof: If $\tp(a/A\overline{b})$, then there is a formula $\phi(x,\overline{b})\in \tp(a/A\overline{b})$ characterizing the type, i.e., the set of realisations of $\tp(a/A\overline{b})$ is exactly the finite set of realisations of the formula $\phi(x,\overline{b})$. Suppose this set is $\{a=a_1,a_2,\ldots,a_n\}$.
By assumption, $\tp(a_i/A)=\tp(a/A)$ is not algebraic for all $i=1,2,\ldots,n$, so by the lemma there is a tuple $(a_1^1,\ldots,a_n^1)\equiv_A (a_1,\ldots,a_n)$ which is disjoint from $B_1=\{a_1,\ldots,a_n\}$. Say $\alpha_1$ is an automorphism over $A$ sending $(a_1,\ldots,a_n)$ to $(a_1',\ldots,a_n')$. Then we have $\overline{b}\equiv_A \alpha_1(\overline{b})$ and $\phi(M,\overline{b})\cap \phi(M,\alpha_1(\overline{b}))=\emptyset$, and we can call $B_2=\{a_1,\ldots,a_n,a_1^1,\ldots,a_n^1\}$
Applying repeatedly the lemma, we can always find a tuple $(a_1^m,\ldots,a_n^m)\equiv_A (a_1,\ldots,a_n)$ disjoint from $B_1\cup B_2\cup \cdots \cup B_m$, and by choosing appropriately $\alpha_m \in \Aut(M/A)$ we have $\overline{b}\equiv_A\alpha_m(\overline{b})$ and $\phi(M,\alpha_i(\overline{b})\cap \phi(M,\alpha_m(\overline{b}))=\emptyset$ for every $i<m$.
Thus, the formula $\phi(x,\overline{b})$ 2-divides over $A$, witnessed by the infinite sequence $\langle \alpha_m(\overline{b}):m<\omega\rangle$. $\Box$
