Formula witnessing Morley Rank.

Let $$T$$ be an $$\omega$$-stable theory, and $$P$$ a strongly minimal, $$\varnothing$$-definable set inside of a saturated model $$M$$. Let $$a \in M$$, $$b \in P$$ such that $$tp(a/b)$$ is $$P$$-internal, that is, $$a \in dcl(d ,p)$$, for some $$d$$ independent from $$a$$ over $$b$$, and some $$p \in P$$. Let $$m = MR (tp(a/b))$$, then there is a formula $$\varphi(x,y)$$ such that $$\models\varphi(a,b)$$ and for every $$b' \in P$$, if $$\varphi(x,b')$$ is consistent, then $$MR(\varphi(x,b'))=m$$.

I am not sure how to construct this formula, as I don't think we can just pick some $$\varphi$$ witnessing the Morley rank of $$tp(a/b)$$, since we need the rank to be preserved as we change the parameters (without leaving $$P$$). I believe it is necessary to use the fact that $$P$$ is strongly minimal in order to be able to express the Morley Rank of a formula as an elementary property of its parameters. This would be possible if $$a \in P$$, but we only have $$a \in dcl(d,P)$$ for some tuple $$d$$ (I don't see how this could be enough). Am I on the right track? This is mentioned vaguely by the author, but I've been so far unable understand the full argument.

Any help is appreciated!

Since $$a\in \mathrm{dcl}(d,p)$$, there is a function $$f\colon P^n\to M^m$$, defined by $$\chi(w,x,d)$$, such that $$f(p) = a$$ (here $$m$$ is the length of the tuples $$a$$ and $$x$$, and $$n$$ is the length of the tuples $$p$$ and $$w$$). For each $$a'\in M$$, the fiber $$P_{a'} = \{p'\in P^n\mid f(p') = a'\}$$ is an $$a'd$$-definable set (uniformly in $$a'$$).

Let $$k = \mathrm{MR}(P_{a})$$. Then by definability of Morley rank in $$P$$, the set $$X = \{a'\in M\mid \mathrm{MR}(P_{a'}) = k\}$$ is a $$d$$-definable set containing $$a$$.

Since $$a$$ and $$d$$ are independent over $$b$$, we have $$\mathrm{MR}(\mathrm{tp}(a/bd)) = \mathrm{MR}(\mathrm{tp}(a/b))=m$$ So there is a $$bd$$-definable set $$X'\subseteq X$$ containing $$a$$ such that $$\mathrm{MR}(X') = m$$.

Let $$\psi(x,b,d)$$ be the formula defining $$X'$$.

Let $$Y = f^{-1}(X')\subseteq P^n$$, and note that $$Y$$ is $$bd$$-definable.

Then $$f$$ is a definable surjection from $$Y$$ onto $$X'$$ such that each fiber has Morley rank $$k$$, and hence $$\mathrm{MR}(Y) = m+k$$.

Now let $$\theta(x,y,z)$$ be the conjunction of formulas expressing:

1. $$\psi(x,y,z)$$.

2. $$\chi(w,x',z)$$ defines a function $$f_z\colon P^n\to M^m$$ whose image contains the set in $$M^m$$ defined by $$\psi(x',y,z)$$.

3. The $$f_z$$-preimage of the set in $$M^m$$ defined by $$\psi(x',y,z)$$ has Morley rank $$m+k$$.

4. For all $$x'$$ such that $$\psi(x',y,z)$$, the fiber of $$f_z$$ over $$x'$$ has rank $$k$$.

For 3 and 4, we again use definability of Morley rank in $$P$$. Note that conditions 2-4 only involve the variables $$y$$ and $$z$$, not $$x$$, so for any $$b'$$ and $$d'$$, $$\theta(x,b',d')$$ is either empty or equivalent to $$\psi(x,b',d')$$.

We certainly have $$\theta(a,b,d)$$. And for any $$b'$$ and $$d'$$, the set defined by $$\theta(x,b',d')$$, if it is non-empty, has Morley rank $$m$$ (since it is the image of a set of rank $$m+k$$ under a definable surjection in which each fiber has rank $$k$$). So $$\exists z\, \theta(x,y,z)$$ does what you want.