1) I was wondering about the following statement (it can be found in Makkai): Assume $T$ is stable. In $T^{eq}$ we have that $A_1\overset{\vert}{\smile}_{A} A_2$ if and only if $acl(A_1)\overset{\vert}{\smile}_{acl{(A)}} acl(A_2)$. I was wondering why $T^{eq}$ is necessary for this. Is there a counterexample for the faliure of this if we just look at a theory $T$ that doesn't eliminate imaginaries?
2) On a related note: suppose that we are in a stable theory (or even simple theory) and so forking = dividing. Is it true that $A_1 \overset{\vert}{\smile}_{A} A_2$ if and only if $A \overset{\vert}{\smile}_{acl(A)} A_2$? Unless there is an error in the following, it should be true:
Since any sequence of indiscernibles $A$ is also indiscernible over $acl(A)$ it follows that a formula that divides over $A$ also divides over $acl(A)$. This proves the forward direction. For the convesrse take an formula $\phi(x,a_1,a_2)$ that divides over $acl(A)$ with $a_1$ being a tuple that contains the algebraic types over $A$ that appear in the formula. Let $\psi(y)$ by an $L(A)$ formula that witnesses the algebracity of $a_1$. We may assume that $\psi(y)$ has the least number of realizations possible. It then follows that $\psi(y)$ isolates the type of $a_1$ over $A$. Now consider the formula $\psi'(x)=\exists{y}(\phi(x,y,a_2)\wedge{\psi(y)})$. Then this formula divides: take a sequence that witnesses that $\phi$ divides. If you assume to the contrary then we $\psi'$ is $k$ consistent for all $k$. But then we can fins an automorphism that fixes $A$ and the indiscernibles sends the witness of $\psi$ to $a_2$ which contradicts our initial assumption.
Is this proof correct am I missing something?