This question assumes some familiarity with basic stability theory. We work in a saturated model of a stable theory eliminating imaginaries. If $a \in \text{dcl}(C,b)$, then $a\in\text{acl}(\text{Cb}(a,b/C ),b)$ by forking calculus. My question is if in fact $a\in\text{dcl}(\text{Cb}(a,b/C ),b)$?
I can prove it in the case when the ambient theory is totally trivial: Suppose the formula $\varphi(x,b,c)$ witnesses that $a \in \text{dcl}(C,b)$. Suppose that $\sigma$ is an automorphism such that $\sigma(\text{Cb}(a,b/C ))=\text{Cb}(a,b/C )$ and $\sigma(b)=b$. We have to show that $\sigma(a)=a$. Note that $p_1=\text{tp}(a,b/C)$ and $p_2=\text{tp}(\sigma(a),b/\sigma(C))$ are parallel, i.e. have a common nonforking extension $p\in S(C\cup \sigma(C))$. Let $(a',b')$ be a realisation of $p$. Note that:
- $b$ is independent from $C$ over $\text{Cb}(a,b/C )$
- $b$ is independent from $\sigma(C)$ over $\text{Cb}(a,b/C )$
Total triviality implies that $b$ is independent from $C \cup \sigma(C)$ over $\text{Cb}(a,b/C )$. Since the type of $b$ over $\text{Cb}(a,b/C )$ is stationary, it follows that $b'$ and $b$ have the same type over $C\cup \sigma(C)$. Hence there is $a''$ such that $(a'',b)$ realises $p$. Then
- $(a'',b)$ realises $p_1$ $\Rightarrow$ $\varphi(a'',b,c)$ $\Rightarrow$ $a''=a$ and $\varphi(\sigma(a''),b,\sigma(c))$
- $(a'',b)$ realises $p_2$ $\Rightarrow$ $\varphi(a'',b,\sigma(c))$
Hence $\sigma(a)=\sigma(a'')=a''=a$.
Assume your assumptions. Take $a\in\text{dcl}(b\mathfrak{C})$. Assume that $a\not\in\text{dcl}(\text{Cb}(a,b/\mathfrak{C}), b)$. Then there exists some $a'$ distinct from $a$ and they both realize the same algebraic formula over $(\text{Cb}(a,b/\mathfrak{C}), b)$.
If $a, a'$ have the same orbit over $(\text{Cb}(a,b/\mathfrak{C}), b)$ and assume that $(\text{Cb}(a,b/\mathfrak{C}), b)$ is small (should hold when T is stable), then there is an automorphism $\sigma$ over $($Cb$(a,b/\mathfrak{C}), b)$ sending $a$ to $a'$. Since $\sigma$ fixes Cb$(a,b/\mathfrak{C})$, it must fix tp$(a,b/\mathfrak{C})$ by the definition of canonical basis, which means that $$\text{tp}(a,b/\mathfrak{C})=\text{tp}(\sigma(a),\sigma(b)/\mathfrak{C})=\text{tp}(a',b/\mathfrak{C}).$$ It follows that tp$(a/b\mathfrak{C})=\text{tp}(a'/b\mathfrak{C})$. By $a\in\text{dcl}(b\mathfrak{C})$, we have $a=a'$.
Assume that $a,a'$ are in different orbits over $(\text{Cb}(a,b/\mathfrak{C}),b)$. Use the fact: for a small set $A\subseteq\mathfrak{C}$, the orbit of $a$ under the action of $\text{Aut}(\mathfrak{C}/A)$ is singleton iff $a\in\text{dcl}(A)$.
Proof of the fact: For the nontrivial direction '$\Rightarrow$' of this fact, note that if $a$ and $a'$ are in different orbit, the proof I knew was to say that tp$(a/A)$ has two realizations (by a small set of formulas), which is inconsistent by saturation, so it results in a formula witnessing $a\in\text{dcl}(A)$.
It was quite surprising when I think about the fact in this direction: why cannot $a,a'$ be realizations of the same algebraic formula, while they both have orbit as singletons. It turned out that saturation really yields a formula defining them separately.