Morley rank (with an unusual definition) For a definable set $X \subseteq \mathbb{U}^n$, let us denote $\text{RM}(X)$ the Morley Rank $\text{RM}(\varphi(\bar{x}))$, with $\varphi(\bar{x})$ the formula defining $X$. 
Show that, for $X \subseteq \mathbb{U}^n$ and $Y \subseteq \mathbb{U}^m$, we have $\text{RM}(X \times Y)= \text{RM}(X) + \text{RM}(Y)$.
I know this should be some kind of standard result. My problem is we defined $\text{RM}$ just in case of theory $T$ strongly minimal as follows: for a type $p(\bar{x}) \in S(A)$, let $RM(p)$ be $\dim(\bar{a}/A)$ for any realization $\bar{a}$ of $p(\bar{x})$. So I suppose I have to use this definition in order to show the result above. 
 A: In the comments, you've defined the Morley rank of a formula in a strongly minimal theory as follows:
$\text{RM}(\varphi(\overline{x})) = \text{max}\{\text{dim}(\overline{a}/A)\mid \mathbb{U}\models\varphi(\overline{a})\}$, where $A\subseteq \mathbb{U}$ is a small set and $\varphi(\overline{x})$ is a formula over $A$.
Of course, a formula $\varphi(\overline{x})$ can be over $A$ for many choices of $A$ (i.e. any $A$ containing the parameters of $\varphi$), so to make sure the definition is well-defined, you have to check that it doesn't depend on $A$. 
I'll use this fact, as well as the following dimension formula:
$$\dim(\overline{a}\overline{b}/A) = \dim(\overline{a}/A) + \dim(\overline{b}/A\overline{a}).$$
If $X$ is defined by $\varphi(\overline{x})$ and $Y$ is defined by $\psi(\overline{y})$, then $X\times Y$ is defined by $\varphi(\overline{x})\land \psi(\overline{y})$. Let $\overline{a}\overline{b}$ of maximal dimension over $A$ realize $\varphi(\overline{x})\land \psi(\overline{y})$. Then, $$\text{RM}(X\times Y)  = \dim(\overline{a}\overline{b}/A) = \dim(\overline{a}/A) + \dim(\overline{b}/A\overline{a}) \leq \dim(\overline{a}/A) + \dim(\overline{b}/A) \leq \text{RM}(X) + \text{RM}(Y).$$
In the other direction, let $\overline{a}$ of maximal dimension over $A$ realize $\varphi(\overline{x})$, and thinking of $\psi(\overline{y})$ as being over $A\overline{a}$ now, let $\overline{b}$ of maximal dimension over $A\overline{a}$ realize $\psi(\overline{y})$. Then, $$\text{RM}(X)+\text{RM}(Y) = \dim(\overline{a}/A) + \dim(\overline{b}/A\overline{a}) = \dim(\overline{a}\overline{b}) \leq \text{RM}(X\times Y).$$
Edit: A proof the dimension formula, as requested. 
Let $\overline{a}'$ be a maximal subtuple of $\overline{a}$ which is independent over $A$, and let $\overline{b}'$ be an maximal independent subtuple of $\overline{b}$ which is independent over $A\overline{a}$ (equivalently, over $A\overline{a}'$). I claim that $\overline{a}'\overline{b}'$ is a maximal subtuple of $\overline{a}\overline{b}$ which is independent over $A$. Since the dimensions in question are the lengths of these subtuples, this is all we need to show.
It's easy to see that $\overline{a}'\overline{b}'$ is maximal. If it weren't, we could add some new $a$ from $\overline{a}$ (but then $\overline{a}'a$ would be independent over $A$, contradicting maximality) or some new $b$ from $\overline{b}$ (but then $\overline{b}'b$ would be independent over $A\overline{a}$, contradicting maximality). 
To see that it's independent, note that if we had some $b\in \overline{b}'$ such that $b\in \text{acl}(A\overline{a}'\overline{b}'')$, where $\overline{b}''$ is $\overline{b}'$ with $b$ removed, then $\overline{b}'$ would not be independent over $A\overline{a}'$. Similarly, if we had some $a\in \overline{a}'$ such that $a\in \text{acl}(A\overline{a}''\overline{b}')$, where $\overline{a}''$ is $\overline{a}'$ with $a$ removed, then we could cut down $\overline{b}'$ to a minimal subtuple $\overline{b}''$ such that $a\in \text{acl}(A\overline{a}''\overline{b}'')$ (and $\overline{b}''$ is not empty since $\overline{a}'$ is independent over $A$), and use exchange to get some $b\in \overline{b}''$ to have $b\in \text{acl}(A\overline{a}'\overline{b}''')$, where $\overline{b}'''$ is $\overline{b}''$ with $b$ removed, contradicting independence of $\overline{b}'$ over $A\overline{a}'$.
That last step actually embedded the key step in for my answer to your previous question. We could make the argument smoother and more symmetrical by noting that symmetry of forking in this context implies that also $\overline{a}'$ is independent over $A\overline{b}'$. 
