How to prove symmetry of forking?

Question

Where can I find a detailed proof for symmetry of forking (page 235 in Marker)? It may be easy but I really can't understand it. For example I don't know why we take $M$ as an $\omega$-saturated model? How in $\omega$-saturated models we can find a formula of rank $t$ which has Morley degree $1$? In which part of proof we use that the Morley degree is $1$?

Answer 1

Just for some background, Marker chose in his book to develop stability theory and forking in the very special case of $\omega$-stable theories. As a result, he defines forking in terms of Morley rank. Forking makes sense in an arbitrary theory $T$, but with a different definition - Marker's definition is only equivalent to the general definition when $T$ is $\omega$-stable.

So I can't point you to another source for this particular proof, as you requested. Of course, there are many good references for symmetry of forking using the general definition, but Marker's book is the main place I'm aware of where the theory of forking is developed just in terms of Morley rank. Anyway, Marker's proof is pretty detailed - it just requires you to have a firm understanding of all the material that came before it in Chapter 6. Let me try to answer your questions.

I don't know why we take $M$ as a $\omega$-saturated model?

The strategy of the proof is to first prove the result in the special case that the set of parameters $A$ is an $\aleph_0$-saturated model $M$ (this is the second paragraph), and then reduce the general case to the special case (this is the third paragraph). As for why we make this reduction, it's because we want to use Lemma 6.4.3, which has the hypothesis that $M$ is an $\aleph_0$-saturated model.

How in $\omega$-saturated models we can find a formula of rank $t$ which has Morley degree $1$?

Here $\omega$-saturation is irrelevant, but it is important that $M$ is a model, in order to get Morley degree $1$. We know that $\text{tp}(a/M)$ has Morley rank $\alpha$. Corollary 6.3.12 tells us that any type over a model has Morley degree $1$. And by the definition of Morley rank and degree of a type (Def. 6.2.12, and note the remark immediately after it), we can find some formula $\varphi(x)\in \text{tp}(a/M)$ with Morley rank $\alpha$ and Morley degree $1$ (and similarly for $\text{tp}(b/M)$).

In which part of proof we use that the Morley degree is $1$?

It seems we don't. It would have been sufficient to take the formula $\varphi$ and $\psi$ to have Morley rank $\alpha$ and $\beta$, respectively, without worrying about their Morley degrees, so we didn't actually need to use that $M$ was a model there. But again, the key point of the proof is the application of Lemma 6.4.3, which requires us to be working over an $\aleph_0$-saturated model.