$MR(tp(a/acl(A))=MR(tp(a/A))$ where MR is morley rank

Question

$MR(tp(a/acl(A))=MR(tp(a/A))$ where MR donotes morley rank. I can't relate $acl(A)$ to $A$. Taking $\varphi(x;a)$ in the $tp(a/acl(A))$ having same rank of the type, I thought there should be some $a'$ conjugate of $a$ over $A$ with $a'\in A$ so that $tp(a)=tp(a')$ and $\varphi(x;a')$ has same rank as $\varphi(x;a)$ but I don't properly do this.

Answer 1

Clearly we have $MR(tp(b/acl(A))) \leq MR(tp(b/A))$, so let us prove $MR(tp(b/A)) \leq MR(tp(b/acl(A)))$.

Let $\varphi(x, a) \in tp(b/acl(A))$ be of minimal rank $\alpha$, where $a$ denotes the parameters from $acl(A)$. Then the type $tp(a/A)$ is algebraic and is thus isolated by some algebraic $L(A)$-formula $\psi(y)$. Denote by $\{a_1, \ldots, a_n\}$ the realisations of $\psi(y)$, then $$ \exists y(\psi(y) \wedge \varphi(x, y)) $$ and $$ \varphi(x, a_1) \vee \ldots \vee \varphi(x, a_n) $$ are equivalent. So $$ MR(\exists y(\psi(y) \wedge \varphi(x, y))) = MR(\varphi(x, a_1) \vee \ldots \vee \varphi(x, a_n)) = \max(MR(\varphi(x, a_1)), \ldots, MR(\varphi(x, a_n))). $$ For every realisation $a_i$ of $\psi(y)$, we have that $\varphi(x, a)$ and $\varphi(x, a_i)$ have the same Morley rank since the Morley rank only depends on the type of the parameters. So we have for all $1 \leq i \leq n$: $$ MR(\varphi(x, a_i)) = MR(\varphi(x, a)) = \alpha. $$ This means that $MR(\exists y(\psi(y) \wedge \varphi(x, y))) = \alpha$. Now the result follows because $\exists y(\psi(y) \wedge \varphi(x, y))$ is an $L(A)$-formula that is in $tp(b/A)$.