In order to prove that in a strongly minimal structure the Morley rank and the dimension are the same thing, I have to prove the following:
Let $X\subset N^k$ a definable set and $N$ a strongly minimal structure (i.e., every definable set is finite or cofinite and this is true also for elementary extensions of $N$). Then $dim(X)\geq r+1$ if and only if there are infinite definable disjoint sets $X_i\subset X$ s.t. $dim(X_i)\geq r$ for all $i$.
Now the implication from left to right is clear to me. What about the converse?
I tried to use the following
$f:X\longrightarrow Y$ definable and surjective, with fibers of constant dimension $k$. Then $dim(X)=k+ dim(Y)$.
in order to apply some kind of induction, but I fail to see what the function can be.
EDIT: The dimension of a $n$-uple $\overline a\in M^n$ over a set $A\subset M$ is defined as the cardinality of a maximal algebraically independent subset (in the model theoretic sense) of the $n$-uple (in particular it is $\leq n$).
The dimension of a set $X\subset M^n$ is defined as the maximum of the dimension of the $n$-uples which belong to $X$.
CONCLUSION: In light of this result, the Morley rank and the dimension satisfy the same axioms (when they are finite) and then they coincide. But why is the result true?
Thank you in advance.