# Confusion on exercise on $(\mathbb Z, +)$ in Buechler's Stability theory book

Given as exercise 2.5.1:

Show that the complete type realized by $$1$$ in $$(\mathbb Z,+)$$ is non-isolated. HINT Use the preceding exercise.

The previous exercise:

Let $$\vec a$$ and $$\vec b$$ be finite sequences from $$\mathcal M.$$ Prove that $$\operatorname{tp}_{\mathcal M}(\vec a\vec b)$$ is isolated iff $$\operatorname{tp}_{\mathcal M}(\vec a/\vec b)$$ and $$\operatorname{tp}_{\mathcal M}(\vec b)$$ are both isolated. Using this fact, show that when $$\mathcal M$$ is an atomic model and $$\vec b\in M,$$ then $$\mathcal M$$ is atomic over $$\vec b.$$ Conversely, if $$\mathcal M$$ is atomic over $$\vec b$$ and $$\operatorname{tp}_{\mathcal M}(\vec b)$$ is isolated, then $$\mathcal M$$ is atomic.

The only way I can see to "use the hint" here is to show $$(\mathbb Z,+)$$ is atomic over $$1$$ (which is clear since every element is definable from $$1$$) and then by the last part of the previous exercise, if $$\operatorname{tp}(1)$$ were isolated, $$(\mathbb Z, +)$$ would be atomic, hence prime, and there is a mention earlier on page 13, without proof, that $$(\mathbb Z, +)$$ is not prime.

When I was writing up solutions a while ago, I noted if indeed this was the intended solution, it was rather artificial, since thinking about it and looking at the reference given on page 13 (A 'natural' theory without a prime model, by Baldwin, Blass, Glass, and Kueker.), it seemed any reasonable way to show $$(\mathbb Z,+)$$ is not prime would basically amount to showing $$\operatorname{tp}(1)$$ is not isolated.

• Well, perhaps you can somehow explicitly construct a model of Presburger which does not contain a copy of $\mathbf Z$. But I agree that it is hard to conceive an argument for that sort of thing which would not include a reasoning about the type of $1$, and so directly yield its not being isolated. On the other hand, since being prime is basically the same as being countable atomic, the fact that all this is vaguely tautological is not that surprising. Apr 19 at 15:37
• @tomasz Yes, exactly. Following either of the links there is an explicit example of a model that doesn't elementarily embed $\mathbb Z$ (though all models embed $\mathbb Z$ non-elementarily, of course), but you show this by showing the model doesn't realize $\operatorname{tp}_{\mathbb Z}(1)$, i.e. every element is divisible by a prime. Unless there is some model where it's e,g, much easier to show $\operatorname{tp}_{\mathbb Z}(3)$ isn't realized, this seems the inevitable pattern. It's not surprising to me either (I would be surprised to get an answer indicating the hint is useful). Apr 25 at 16:04