# Exercises involving Morley rank & degree

The definitions of Morley rank & degree I use are  I understand these definitions, but I am having a hard time to use them concretely in exercises. For example,

Let $$L$$ be a countable language and $$A$$ an $$\omega$$-saturated $$L$$-structure.

(a) let $$p$$ be a ranked type over $$A$$. Show that $$p$$ has Morley degree 1.

(b) Let $$\varphi(x_1, \dots, x_n)$$ be a ranked $$L_A$$-formula, show that

$$\text{dM}(\varphi(\bar{x})) = |\{p \in S_n(\text{Th}_{L_A}(A)) : \varphi \in p \text{ and } \text{RM}(p) = \text{RM}(\varphi) \}|$$

I don't know how to connect $$\omega$$-saturation with Morley rank & degree. I thought to use that $$\text{Th}(A)$$ has an $$\omega$$-saturated model iff its type spaces are countable iff there is no binary tree of consistent formulas, in the hope that $$\text{dM}(p) \geq 2$$ would allow me to construct such a binary tree, but I cannot get it to work.

• (1) "I don't know how to connect $\omega$-saturation with Morley rank & degree". The connection is that $\omega$-saturation is assumed in your definitions of Morley rank and degree. So you can't even talk about Morley rank and degree (using your definition) over models that aren't $\omega$-saturated. (2) "I thought to use that $\text{Th}(A)$ has an $\omega$-saturated model iff its type spaces are countable..." This is not true! Every theory has $\omega$-saturated models. The theorem is: $T$ has a countable $\omega$-saturated model if and only if its type spaces are countable. Jun 17, 2019 at 16:33
• @AlexKruckman (1) Yes, but I am not sure why the $\omega$-saturated condition is even required, unless you want to talk about the Morley rank of a theory (RM of $x=x$). (2) Oh, yes of course. And we were not given that $A$ is countable. So then I am even more stuck. Jun 17, 2019 at 20:46
• The reason for the $\omega$-saturated condition is to ensure the desirable property the Morley rank of a formula only depends on the theory, not the model $A$. Indeed, you can prove that if $M\preceq N$ are both $\omega$-saturated models, and $\varphi(x)$ is an $L_M$-formula, then $\text{RM}_x(M,\varphi(x)) = \text{RM}_x(N,\varphi(x))$. If you evaluate Morley rank using your definition in a non-$\omega$-saturated model, you may get a smaller value. Jun 18, 2019 at 13:33

I'm assuming your source defines Morley degree to be the $$k$$ from Lemma 5.10.
For (a), Lemma 5.10 ensures that every $$L_A$$-formula $$\varphi(x)$$ can be decomposed into $$k$$ disjoint pieces of full Morley rank, where $$k = \text{dM}(\varphi(x))$$. The main observation is that the formulas $$\varphi_j(x)$$ in this maximal decomposition all have Morley degree $$1$$. Indeed, if $$\varphi_j(x)$$ had Morley degree $$d>1$$, then applying the lemma to this formula, we could split $$\varphi_j(x)$$ into $$d$$ disjoint pieces of full Morley rank. Replacing $$\varphi_j(x)$$ with these formulas in the original decomposition of $$\varphi(x)$$ would decompose $$\varphi(x)$$ into $$k-1+d > k$$ disjoint pieces of full Morley rank.
Now for any complete type $$p(x)$$, let $$\varphi(x)$$ be a formula in $$p(x)$$ of minimal Morley rank and degree. Suppose for contradiction that $$\text{dM}(\varphi(x)) = k > 1$$. Then $$\varphi(x) \leftrightarrow \bigvee_{j=1}^k \varphi_j(x)$$, and each $$\varphi_j(x)$$ has Morley rank equal to $$\text{RM}(\varphi(x))$$ and Morley degree $$1$$. Since $$p(x)$$ is a complete type, $$\varphi_j(x)\in p(x)$$ for some $$1\leq j\leq k$$, contradicting minimality of $$\varphi(x)$$.
For (b), I'll give you a hint. Let $$k = \text{dM}(\varphi(x))$$, and let $$n$$ be the number of complete types of full Morley rank containing $$\varphi(x)$$. You can prove $$k\leq n$$ by decomposing $$\varphi(x)$$ by Lemma 5.10 and extending each formula in the decomposition to a complete type of maximum Morley rank. And you can prove $$n\leq k$$ by picking disjoint formulas containing each of the types of full Morley rank, and applying the bound from Lemma 5.10.
• I think I have (b): for $k \leq n$ you use that $\text{RM}(\varphi(x) \lor \psi(x)) = \max(\text{RM}(\varphi(x)), \text{RM}(\psi(x)))$ and $\text{RM}(x=x) = \text{RM}(\varphi(x) \lor \neg\varphi(x)) \geq \text{RM}(\chi(x))$ for any $\chi(x)$, to extend $\varphi, \varphi_i$ to a complete type $p$ such that $\text{RM}(p) = \text{RM}(\varphi)$. These types must be different because over $\text{Th}_A(A)$ the $\varphi_i$ are incompatible. Jun 20, 2019 at 10:40
• For $n \leq k$: assume WLOG that $n < \infty$, there are $\psi_i$ such that $[\psi_i \land \varphi] \cap \{p_j\}_{j \leq n} = \{p_i\}$. Also $\text{RM}(\psi_i \land \varphi) = \text{RM}(\varphi)$, so the formulas $\psi_i \land \varphi$ witness that $k = dM(\varphi) \geq n$. Jun 20, 2019 at 10:52