Morley Rank of the cartesian product Let $X$ and $Y$ be two definable sets with $RM(X), RM(Y)<\omega$.
Can we say something about $RM(X\times Y)$? If $X$ and $Y$ have Morley Rank $0$, then so does $X\times Y$. I think I can prove that $RM(X\times Y)\geq RM(X) + RM(Y)$.
Does the equality hold? If it does not, does it hold in any special case? (like $Y=X$ or $X=Y=\mathfrak{C})$
 A: You're correct that $\mathrm{RM}(X\times Y)\geq \mathrm{RM}(X)+\mathrm{RM}(Y)$ in general.
Actually, in the possibly infinite rank case, we get: $\mathrm{RM}(X\times Y) \geq \mathrm{RM}(X) \oplus \mathrm{RM}(Y)$, where $\oplus$ denotes the natural sum of ordinals. Natural sum agrees with ordinary sum on finite ordinals, but it is greater than the ordinary sum in general. For example, $1+\omega = \omega$, but $1\oplus \omega = \omega + 1$. So, for example, if $\mathrm{RM}(X) = 1$ and $\mathrm{RM}(Y) = \omega$, then $$\mathrm{RM}(X\times Y) \geq 1\oplus \omega = \omega+1 > \mathrm{RM}(X) + \mathrm{RM}(Y).$$
Ok, but you asked about the finite rank case. Unfortunately, the equality  $\mathrm{RM}(X\times Y) = \mathrm{RM}(X) + \mathrm{RM}(Y)$ is not true in general, but I don't know a simple counterexample. In the paper Singular properties of Morley rank, Lachlan gives an example (on pp. 151-152) of a definable set $X$ with $\mathrm{RM}(X) = 3$ but $\mathrm{RM}(X\times X) = 7$. It is too involved for me to reproduce here. Lachlan also produces upper and lower bounds for how much the Morley rank can grow in products of definable sets.
One fairly clean result is: When $\mathrm{RM}(X) = 0$, $\mathrm{RM}(X\times Y) = \mathrm{RM}(Y)$, and when $\mathrm{RM}(X)>0$, $$\mathrm{RM}(X\times Y) \leq (\mathrm{RM}(X)\cdot \mathrm{RM}(Y)) + \mathrm{RM}(X) = \mathrm{RM}(X)\cdot (\mathrm{RM}(Y)+1).$$ This is  a consequence of Exercise 6.4.4 in Tent and Ziegler's book A Course in Model Theory.
Ok, what about situations where the equality does hold? Well, it holds whenever the Morley rank is equal to the Lascar rank (see also the last section of Lachlan's paper linked above). This is the case in uncountably categorical theories and in groups of finite Morley rank. The additivity of the Lascar rank (made precise by the "Lascar inequalities") is one of the main reasons the Lascar rank is a key tool in stability theory, even when both Lascar rank and Morley rank are defined.
