Indiscernibles over a model

Question

Working within the framework of a monster model, I wish to show that:

(*) If $(a_{i}:i<{\lambda})$ is an indiscernible sequence over $A$, then there is a model $M$ containing $A$ such that $(a_{i}:i<{\lambda})$ is indiscernible over $M$.

(This is en route to establishing that a sequence of indiscernibles over $A$ is also indiscernible over $acl(A)$, which is immediate from the above as $acl(A)\subseteq{M}$ for any $M\models{T}$).

I'm not really sure how to prove the above statement and I'm also curious to know if the second statement can be proved without going through (*) by a simple compactness argument.

My attempt to do so went as follows:

(**) Assume that the sequence is not indiscernible over $acl(A)$. Then there is an $L(acl(A))$ formula $\varphi{(x,\overline{a'})}$ s.t. $\varphi{(x,\overline{a'})}\in{tp(a_{i}/A)}$ but $\neg\varphi{(x,\overline{a'})}\in{tp(a_{j}/A)}$. Since $\overline{a}=a'_0...a'_n$ and each $a'_{i}$ is algebraic I can find formulas $\psi_{a'_{i}}(x)$ over $A$ that are algebraic. My idea then was to use some combination of the $\psi_{a'_{i}}$ with the $\overline{a'}$ being quantified out (something like $\forall{y}$ $(\psi_{a'_i}(y)\wedge{\varphi{(x,y)}}$) (in the case the length of $\overline{a'}$ is 1) and use this formula to show that this witnesses a difference between the types over $A$).

However I can't get this idea to work out, mainly because the algebraic type may have more than one realization. This also has the drawback that the argument would be more complicated if the failure of indiscernibility is witnessed by a finite sequence of elements as opposed to just one. But I'm not certain if this isn't fixable.

To sum up:

1) How do you prove (*)?

2) Is the idea in (**) viable?

Answer 1

I haven't thought too much about your proof approach, but there is a fairly straightforward, entirely different proof, which I'll present here. The proof uses the following standard lemma about the existence of indiscernible sequences. (I think Tent and Ziegler's book actually calls it the "Standard Lemma.") It follows from the usual compactness + Ramsey's theorem method for constructing indiscernible sequences.

Lemma. Let $I, J$ be infinite ordered sets, $\{a_i\}_{i \in I}$ be a sequence, and $A$ any parameter set. Set $\Gamma_n(x_1,\ldots,x_n)$ to be the partial type consisting of those formulas over $A$ which are true of every in-order $n$-tuple from $\{a_i\}$. Then there is an $A$-indiscernible sequence $\{a'_j\}_{j \in J}$ such that, for each $n$, every in-order $n$-tuple from $\{a'_j\}$ satisfies $\Gamma_n(x_1,\ldots,x_n)$.

In particular, if $I = J$ and $\{a_i\}$ is already indiscernible over $A_0 \subseteq A$, then $\{a'_i\} \equiv_{A_0} \{a_i\}$, since $\Gamma_n$ will imply a complete $n$-type over $A_0$.

In our case, take $M'$ any model containing $A$. Then we can use the above lemma to build $\{a'_i\}$ indiscernible over $M'$, with $\{a'_i\} \equiv_A \{a_i\}$. By homogeneity of the monster model, take $f$ an automorphism over $A$ sending $\{a'_i\}$ to $\{a_i\}$, and set $M = f(M')$. Then we have that our original $\{a_i\}$ is $M$-indiscernible, as required.