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!


1 Answer 1


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.


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