# Atomic models of countable complete theories (without finite models) posessing a prime model

I would very much appreciate a hint as I have currently found myself stuck with exercise 13.2.2 in 'Introduction to Model Theory' by Philipp Rothmaler. Before relaying the problem I should perhaps also point out that in this text all theories are defined to be consistent as well as deductively closed.

The problem is as follows:

Suppose $$T$$ is a countable complete theory without finite models. Show that if $$T$$ has an (elementarily) prime model which is not minimal then $$T$$ has an atomic model of power $$\aleph_1$$.

Certainly, if $$\mathfrak{N}$$ is a prime model of $$T$$ then $$\mathfrak{N}$$ is unique up to isomorphism as well as atomic (and non-minimal by assumption). I guess the idea is to judiciously construct an uncountable atomic model out of countable atomic models and making use of the downwards Löwenheim-Skolem theorem but I am certainly missing something.

Hint: Build an elementary chain $$(M_\alpha)_{\alpha<\aleph_1}$$ of countable models, with $$M_\alpha$$ a proper elementary substructure of $$M_\beta$$ for all $$\alpha<\beta$$, and such that each $$M_\alpha$$ is isomorphic to the prime model $$M$$. Show that the union of this chain is atomic and has cardinality $$\aleph_1$$.

To handle the limit step of the transfinite construction, you will need to use the fact that a countable model is atomic if and only if it is prime, and countable prime models are unique up to isomorphism if they exist.

• Dangit, beat me to it! Jan 25, 2020 at 20:21
• @NoahSchweber Well, the OP only asked for a hint - and I was able to give write a hint quickly than your more complete answer. :0) Jan 25, 2020 at 20:22
• Quick question: do we actually need that countable prime models are unique? It seems to me that it's enough to observe that if one countable prime model is non-minimal, then they all are (which is basically trivial). Jan 25, 2020 at 20:44
• @NoahSchweber You're right, that observation suffices. I guess it depends on whether you prefer to take one line to cite a nontrivial result you already know or two lines to explain a trivial observation, heh. Jan 25, 2020 at 20:56

Suppose $$\mathfrak{N}$$ is a non-minimal prime model of its theory. Then there is a proper elementary substructure of $$\mathfrak{N}$$ which is isomorphic to $$\mathfrak{N}$$.

Turning this around, we get: if $$\mathfrak{A}\cong\mathfrak{N}$$ then there is a proper elementary extension of $$\mathfrak{A}$$ also isomorphic to $$\mathfrak{N}$$.

The naive idea at this point is to iterate this: build an increasing elementary chain of copies of $$\mathfrak{N}$$ of length $$\omega_1$$. Every element of the union will have its type determined (by elementarity) by the model it appears in, and since all the models in the chain are atomic that type will be principal. So the result will be atomic (and clearly of size $$\aleph_1$$).

However, it's not that simple: consider what happens at "stage $$\omega$$" of the chain. We have $$\mathfrak{A}_0\prec\mathfrak{A}_1\prec\mathfrak{A}_2\prec ...$$ each isomorphic to $$\mathfrak{N}$$, and we now need to argue that $$\bigcup_{i\in\omega}\mathfrak{A}_i$$ is, or can be elementarily embedded in a structure which is, isomorphic to $$\mathfrak{N}$$ again. And in fact there has to be a subtlety here, since the exercise is not true if we replace $$\aleph_1$$ with $$\aleph_2$$ ...