Problem with forking

Question

Let $A \subseteq \mathbb{U}$ be a small set, and $\bar{a}= (a_1, \dots, a_n)$, $\bar{b}=(b_1, \dots, b_k)$ be tuples of $\mathbb(U)$. Show that tp$(\bar{a}/A\cup (b_1, \dots, b_k)$ does not fork over $A$ if and only if tp$(\bar{b}/A\cup (a_1, \dots, a_n)$ does not fork over $A$.

Answer 1

This proposition is true in much greater generality (all simple theories), but based on your other question and your reference to the exchange property of algebraic closure in the comments, I will assume that you're working with a strongly minimal theory $T$ and the following definition of Morley rank:

For a type $p(\overline{x})∈S(A)$, $\text{RM}(p) = \text{dim}(\overline{a}/A)$ for any realization $\overline{a}$ of $p$, where $\text{dim}$ is the dimension of the $\text{acl}$ pregeometry induced by strong minimality.

First, observe that it suffices to consider the case where $\overline{a}$ and $\overline{b}$ are both algebraically independent tuples over $A$. Indeed, we can refine $\overline{a}$ to an independent subtuple $\overline{a}'$ so $\text{dim}(\overline{a}/A) = \text{dim}(\overline{a}'/A)$, and similarly for $\overline{b}$.

Now we are assuming that $\overline{a}$ is independent over $A$, $\overline{b}$ is independent over $A$, and $\dim(\overline{a}/A) = \dim(\overline{a}/A\overline{b})$, i.e. $\overline{a}$ is independent over $A\cup\{b_1,\dots,b_k\}$, and we would like to show that $\dim(\overline{b}/A) = \dim(\overline{b}/A\overline{a})$. If not, then some $b_i$ (say WLOG it's $b_k$) is in $\text{acl}(A\cup\{a_1,\dots,a_l\}\cup \{b_1,\dots,b_{k-1}\})$.

Let $\mathfrak{a} = \{a_{i_1},\dots,a_{i_j}\}$ be a minimal subset of $\{a_1,\dots,a_l\}$ such that $b_k\in \text{acl}(A\cup\{a_{i_1},\dots,a_{i_j}\}\cup \{b_1,\dots,b_{k-1}\})$. Since $b_k\notin \text{acl}(A\cup\{b_1,\dots,b_{k-1})$, $\mathfrak{a}$ is not empty.

Now we're set up to use exchange: \begin{align*}b_k &\in \text{acl}(A\cup\{a_{i_1},\dots,a_{i_j}\}\cup \{b_1,\dots,b_{k-1}\}),\text{but} \\ b_k &\notin \text{acl}(A\cup\{a_{i_1},\dots,a_{i_{j-1}}\}\cup \{b_1,\dots,b_{k-1}\}),\text{so} \\ a_{i_j} &\in \text{acl}(A\cup\{a_{i_1},\dots,a_{i_{j-1}}\}\cup\{b_1,\dots,b_k\}) \end{align*}

Contradicting the independence of $\overline{a}$ over $A\overline{b}$.