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.

Then I saw this answer today which reminded me about this problem and reinforced my previous take on it.

My question is just if I am making an oversight here... is there something in the hint that's actually useful for seeing the big picture?

  • 1
    $\begingroup$ 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. $\endgroup$
    – tomasz
    Apr 19 at 15:37
  • $\begingroup$ @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). $\endgroup$ Apr 25 at 16:04
  • $\begingroup$ Well, it's very easy to show that type of 1 is realised if and only if the type of 3 (or any other nonzero integer) is realised, since they are all easily interdefinable, so the hypothetical you mention can't really happen. $\endgroup$
    – tomasz
    Apr 25 at 20:14
  • $\begingroup$ @tomasz Oops, missed that... anyway was just grasping for straws with that one. $\endgroup$ Apr 25 at 21:54


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