TITLE: Independence and algebraic closure for stable theories.
QUESTION [6 upvotes]: Let $T$ be a stable theory, and $M$ a sufficiently saturated model. Let $x,b,E $ such that
$x \in \operatorname{acl}(MEb)$ and such that $x\overset{\vert}{\smile}_{Mb}E$. Then $x \in \operatorname{acl}(Mb)$.
On the intuitive level, this is very clear to me, it just says that the $E$ part making $x$ algebraic is unnecessary (by independence), however I can't seem to prove this using the ''axiomatic'' properties of the independence relation $\overset{\vert}{\smile}_{}$ i.e symmety, transitivity, etc. I also tried arguing by contradiction, using the exchange property for $\operatorname{acl}$.
I stumbled upon this small step while working through a proof of the group configuration for stable theories, but I don't think the context matters much.

REPLY [4 votes]: Since you're happy to assume $T$ is stable, the other answers have surely settled your question. But the fact is true in an arbitrary theory $T$, so I'd like to give you an additional proof that does not rely on stability.
First a definition. We say $A$ and $B$ are algebraically independent over $C$, written
$\newcommand{\acl}{\mathrm{acl}}
\newcommand{\tp}{\mathrm{tp}}
\newcommand{\ind}{\overset{\vert}{\smile}} A\ind^a_CB$, if $\acl(AC)\cap \acl(BC) = \acl(C)$. To avoid confusion, I'll write $\ind^f$ for forking independence.
Lemma (Full existence for algebraic independence): Let $A$, $B$, $C$ be arbitrary sets. Then there exists $A'$ with $\tp(A'/C) = \tp(A/C)$ and $A'\ind^a_C B$.
The lemma follows from "Strong elementary amalgamation over algebraically closed sets", Theorem 6.4.5 in Hodges Model Theory or Theorem 5.3.5 in Hodges A Shorter Model Theory.
Now suppose $a\ind^f_B E$ and $a \in \acl(BE)$. Then there is some algebraic $L(B)$-formula $\varphi(x,e)\in \tp(a/BE)$, and we may assume that $\varphi(x,e)$ isolates $\tp(a/BE)$. By the lemma, we can find a sequence $(e_n)_{n\in \omega}$ such that $e_0 = e$, $\tp(e_n/B) = \tp(e/B)$, and $e_n\ind^a_B e_0\dots e_{n-1}$ for all $n$. Since $a\ind^f_BE$, $\varphi(x,e)$ does not divide over $B$. So $\{\varphi(x,e_n)\mid n\in \omega\}$ is not $2$-inconsistent. In particular, there exist $n< n'$ and $a'$ such that $\varphi(a',e_n)$ and $\varphi(a',e_{n'})$. But then $a'\in \acl(Be_{n'})\cap \acl(Be_{n}) = \acl(B)$, since $e_{n'}\ind^a_B e_n$. Finally, since $\tp(a'e_n/B) = \tp(ae/B)$, there is an automorphism of the monster model fixing $B$ and moving $a'e_n$ to $ae$. This automorphism fixes $\acl(B)$ setwise, so $a\in \acl(B)$.