How to show than an algebraic formula divides? Let $A \subseteq B \subseteq M$ be subsets of some model $M$ of a first-order theory. And assume that $tp(c/ A)$ is not algebraic while $tp(c/ B)$ is algebraic. It is clear that $tp(c/ B)$ forks, but I guess one can also show that the formula $\phi(\bar{x},\bar{b})$ that wtinesses the algebraicity of $\bar{c}$ over $B$ actually divides over $A$. However, I have struggle proving such a statement. I assume it is very easy but I can't really see how to prove it. 
 A: One has to be a bit careful with this statement. For example, suppose $A = \{a\}$, $B = \{a,b\}$, and $c = b\notin \text{acl}(A)$. Then the formula $(x = a \lor x = b)$ witnesses that $\text{tp}(c/B)$ is algebraic, but it does not divide over $A$. But if we consider a formula isolating $\text{tp}(c/B)$ (or at least implying that $c\notin \text{acl}(A)$), this formula will divide over $A$. 
One way to prove this is to use P.M. Neumann's lemma (see Corollary 4.2.2 in Model Theory by Hodges, and Exercise 5.6.2 in A Course in Model Theory by Tent & Zielger). I'll use the statement in Hodges. 

Let $A$ be an $L$-structure, and suppose $a$ is a non-empty tuple of elements of $A$ such that no element in $a$ lies in a finite orbit of $\text{Aut}(A)$. Then the orbit of $a$ under $\text{Aut}(A)$ contains an infinite set of pairwise disjoint tuples.

The proof of this statement is an application of B.H. Neumann's lemma, an equivalent purely group-theoretic statement. (My understanding is that B.H. and P.M. Neumann are father and son!)

Ok, now let's be clear about our setting. We have $A\subseteq B$, and an element $c\in \text{acl}(B)\setminus \text{acl}(A)$. Let $\varphi(x,b)\in \text{tp}(c/B)$ be an formula with only finitely many solutions, none of which is in $\text{acl}(A)$ (a formula isolating $\text{tp}(c/B)$ will do). Our goal is to show that $\varphi(x,b)$ divides over $A$. 
Let $c^1,\dots,c^k$ enumerate the finitely many solutions of $\varphi(x,b)$. We also divide the tuple of parameters $b$ into two subtuples: $d$ consists of those elements of $b$ not in $\text{acl}(A)$, and $e$ consists of those elements of $b$ in $\text{acl}(A)$. We apply P.M. Neumann's lemma to (1) a sufficiently saturated model $M_{\text{acl}(A)}$ containing all relevant sets and tuples, with the elements of $\text{acl}(A)$ named by constants, and (2) the tuple $c^1\dots c^k d$. Note that no element of this tuple is in $\text{acl}(A)$, hence no element lies in a finite orbit of $\text{Aut}(M_{\text{acl}(A)})$. 
From the lemma, we get a sequence $(c^1_i\dots c^k_id_i)_{i\in \omega}$ of disjoint tuples which all satisfy the same type over $\text{acl}(A)$ as $c^1\dots c^k d$. This implies that $(c^1_i\dots c^k_id_ie)_{i\in \omega}$ all satisfy the same type over $A$ as $c^1\dots c^k de$. In particular, letting $b_i = d_ie$, the sequence $(b_i)_{i\in \omega}$ witnesses the dividing of $\varphi(x,b)$ over $A$, since the family $\{\varphi(x,b_i)\mid i\in \omega\}$ is pairwise inconsistent: the solutions to $\varphi(x,b_i)$ and $\varphi(x,b_j)$ are the disjoint sets $\{c^1_i,\dots,c^k_i\}$ and $\{c^1_j,\dots,c^k_j\}$. 
