Properties of the types of elements in the algebraic closure. Let $\mathcal{M}$ be a model which is saturated enough ( and strongly homogeneous and everything that is needed).
Consider some element $a$ and the type $p:=tp(a/\mathcal{M})$. Let $b\in acl(a)$. I am trying to understand what properties of $p$ will pass to $tp(b/\mathcal{M})$.
I can prove that if $p$ is $A$ invariant for some $A\subset\mathcal{M}$, then $tp(b/\mathcal{M})$ is also invariant (not over $A$ in general, but over a bigger set of parameters).
It is also reasonable to expect that definability will also pass to $tp(b/\mathcal{M})$, even though I cannot think about an immediate proof.
Are there other properties (at least elementary ones) that pass from p to $tp(b/\mathcal{M})$? Is there any reference I could check?
 A: Here are some examples to get the ball rolling, none of which requires $M$ to be a model; I am sure there is much more to be said. Let $C$ be a parameter set, and let $p=\operatorname{tp}(a/C)$ and $q=\operatorname{tp}(b/C)$. Also, let $\phi(x,y)$ be an $L$-formula such that $\models\phi(a,b)$ and $\phi(a,y)$ is algebraic, say of degree $n$. Let $\widehat{\phi}(x,y)\equiv \phi(x,y)\wedge\exists^{\leqslant n}y\phi(x,y)$.
Finite satisfiability: Suppose $p$ is finitely satisfiable in a set $A$; then $q$ is satisfiable in $\operatorname{acl}A$. Indeed, let $\theta(y)$ be a formula in $q$. The formula $\exists y\left[\widehat{\phi}(x,y)\wedge\theta(y)\right]$ lies in $p$, and is hence satisfiable in $A$, say by $a'$; then there is $b'$ realizing $\widehat{\phi}(a',y)\wedge\theta(y)$. So $b'$ lies in $\operatorname{acl}(a')$, hence in $\operatorname{acl}(A)$, and satisfies $\theta(y)$, as desired.
Definability: Suppose $p$ is definable over a set $A$; then $q$ is definable over $A$. To see this, let $\psi(x,y)$ be an $L_C$-formula such that $\models\psi(a,b)$ and $\psi(a,y)$ is algebraic of minimal degree possible. Then in particular any two realizations of $\psi(a,y)$ have the same type over $C$; call this fact $\star$. Now, let $\theta(y,z)$ be an $L$-formula. Applying definability of $p$ to the $L_C$-formula $\widehat{\theta}(x,z)\equiv\forall y\left[\psi(x,y)\to\theta(y,z)\right]$, there exists an $A$-formula $\xi(z)$ such that, for any tuple $c\in C^{|z|}$, we have $\models\xi(c)$ iff $\models\widehat{\theta}(a,c)$. On the other hand, by $\star$, we also have $\models\theta(b,c)$ if and only if $\models\widehat{\theta}(a,c)$ for any tuple $c\in C^{|z|}$. So $\xi(z)$ gives the desired definition for the $\theta(y,z)$-part of $q$, as desired.
Non-dividing: Suppose $p$ does not divide over a set $A$; then $q$ does not divide over $A$. Indeed, suppose that $\theta(y,c)$ in $q$ divides over $A$, witnessed by an $A$-indiscernible sequence $(c_i)_{i\in\omega}$. Then the formula $\widehat{\theta}(x,c)\equiv\exists y\left[\widehat{\phi}(x,y)\wedge\theta(y,c)\right]$ lies in $p$, and I claim it divides over $a$, witnessed by the same $c_i$. Indeed, if $a'$ realizes $\{\widehat{\theta}(x,c_i)\}_{i\in\omega}$, then (i) the set $B=\{b':\models\phi(a',b')\}$ has size at most $n$, and (ii) for every $i\in\omega$ there is $b'\in B$ with $\models\theta(b',c_i)$. In particular, by the pigeonhole principle, there is an infinite subset $I_0\subseteq\omega$ and a $b'\in B$ such that $\models\theta(b',c_i)$ for every $i\in I_0$, contradicting that the family $\{\theta(y,c_i)\}_{i\in\omega}$ must be $k$-inconsistent for some $k\in\omega$.
Non-forking: Suppose $p$ does not fork over a set $A$; then $q$ does not fork over $A$. For suppose otherwise; then there exists a tuple $e$ such that every extension of $q$ to $Ce$ divides over $A$. Now, $p$ has an extension $p'$ to $Ce$ that does not fork over $A$; let $a'$ be a realization, let $\alpha$ be an automorphism of the monster model fixing $C$ pointwise and taking $a$ to $a'$, and let $b'=\alpha(b)$. Then $b'\in\operatorname{acl}(a')$, and $p'=\operatorname{tp}(a'/Ce)$ does not divide (indeed, does not fork) over $A$, so by the step above $q':=\operatorname{tp}(b'/Ce)$ also does not divide over $A$. But $q'\supseteq q$ is an extension of $q$ to $Ce$, so this is a contradiction.
