Constructing an indescernible sequence from a coheir

Question

What follows is part of Exercise 1.34 from Pillay's Introduction to Stability Theory. Suppose the following:

1. $M \prec N$.
2. $N$ is $|M|^+$-saturated.
3. $p \in S_1(M)$, $q \in S_1(N)$.
4. $q \supset p$ is a coheir of $p$.

Construct a sequence $(a_i \mid i < \omega)$ of elements in $N$ inductively as follows: let $a_0$ realize $p$, and given $a_0, \dots, a_n$, let $a_{n+1}$ realize $q \upharpoonright Ma_0\dots a_n$. The exercise is to show that $(a_i)_i$ is an indiscernible sequence over $M$.

My attempt is to show, by induction on $n$, that for $\bar ab$ and $\bar a'b'$, subsequences of $(a_i)_i$ of length $n$, the type of $\bar a$ over $M$ is the same as that of $\bar a'$. Suppose that the following are equivalent for a formula $\phi$ without parameters, $\bar cd $ a subsequence of $(a_i)_i$, and $\bar m \in M$:

1. $N \models \phi(\bar c, d; \bar m)$
2. There exists $d' \in M$ such that $N \models \phi(\bar c, d';\bar m)$.

Note that since $q$ is a coheir, 1 implies 2. If that is true, what I need to prove is a straightforward consequence of the induction hypothesis.

My question is whether or not the two conditions are equivalent, and how to show it.

Answer 1

There's no hope that your condition $2$ implies your condition $1$. For example, imagine that $\varphi$ has just one free variable, so the implication reduces to "if there exists $d'\in M$ such that $N\models \varphi(d')$, then $N\models \varphi(d)$." Unless $T$ proves that $\varphi$ is true of everything or false of everything, there are going to be elements of $M$ that satisfy $\varphi$ and elements of $M$ that satisfy $\lnot\varphi$, so if $2$ implies $1$, $d$ has to satisfy both $\varphi$ and $\lnot\varphi$.

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I'll spell out the approach hinted at by Levon's answer and by me in the comments to Levon's answer. The key property of coheirs is that they are invariant types.

We say that $q(x)\in S_1(N)$ is $M$-invariant if for every formula $\psi(x,y)$ and all tuples $b, b'\in N$, if $\mathrm{tp}(b/M) = \mathrm{tp}(b'/M)$, then $\psi(x,b)\in q(x)\iff \psi(x,b')\in q(x)$.

Claim: If $q(x)$ is a coheir of its restriction to $M$, then it is $M$-invariant.

Proof: Suppose not. Then there is a formula $\psi(x,y)$ and tuples $b,b'\in N$ such that $\mathrm{tp}(b/M) = \mathrm{tp}(b'/M)$, $\psi(x,b)\in q(x)$, and $\psi(x,b')\notin q(x)$. Hence $\psi(x,b)\land \lnot\psi(x,b')\in q(x)$. But since $q(x)$ is a coheir of its restriction to $M$, there exists $a\in M$ such that $N\models \psi(a,b)\land \lnot\psi(a,b')$. This contradicts our assumption that $\mathrm{tp}(b/M) = \mathrm{tp}(b'/M)$. $\square$

Now forget the fact that $q$ is a coheir and just remember that it's $M$-invariant. Build a sequence as described in your question (this is called a Morley sequence for the invariant type $q$). To show that it's $M$-indiscernible, show by induction on $n$ that $\mathrm{tp}(\overline{a}b/M) = \mathrm{tp}(\overline{a}'b'/M)$ whenever $\overline{a}b$ and $\overline{a}'b'$ are subsequences of length $n+1$. And note that this amounts to showing that for any formula $\psi$, $\psi(x,\overline{a},\overline{m})\in \mathrm{tp}(b/M\overline{a}) = q\restriction M\overline{a}$ if and only if $\psi(x,\overline{a}',\overline{m})\in \mathrm{tp}(b'/M\overline{a}') = q\restriction M\overline{a}'$.

Answer 2

I don't think 2 implies 1 as stated.

My hint for the exercise would be the following: let $\phi(\bar x, y)$ be a formula over $M$ and let $\bar a'' \in N$ realise the type $tp(\bar a/M) = tp(\bar a'/M)$. Show that $\phi(\bar a, b)$ iff $\phi(\bar a'', x) \in q$ iff $\phi(\bar a', b')$.