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The definitions of Morley rank & degree I use are

Morley rank definition

Morley degree definition

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.

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  • $\begingroup$ (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. $\endgroup$
    – Alex Kruckman
    Jun 17, 2019 at 16:33
  • $\begingroup$ @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. $\endgroup$
    – Ibrahim
    Jun 17, 2019 at 20:46
  • $\begingroup$ 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. $\endgroup$
    – Alex Kruckman
    Jun 18, 2019 at 13:33

1 Answer 1

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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.

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  • $\begingroup$ I understand your answer for (a), but what do you mean by "full Morley rank"? $\endgroup$
    – Ibrahim
    Jun 19, 2019 at 11:00
  • $\begingroup$ Equal to the formula in question (not less). $\endgroup$
    – Alex Kruckman
    Jun 19, 2019 at 13:16
  • $\begingroup$ 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. $\endgroup$
    – Ibrahim
    Jun 20, 2019 at 10:40
  • $\begingroup$ 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 $k = dM(\varphi) \geq n$. $\endgroup$
    – Ibrahim
    Jun 20, 2019 at 10:52

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