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I have a hard time proving that the theory of $(\mathbb{Z},+,0)$ does not have any prime model.

Specifically, I need to show that the model $(\mathbb{Z},+,0)$ is not atomic(that's my main goal), but it's equivalent to the fact that for some $0\neq z\in \mathbb{Z}$ the type $tp(z)$ isn't isolated. How can I prove that?(That the type of some element isn't isolated)

I've seen that $(\mathbb{Z},+,0)$ is elementary minimal if it helps.

Eitan
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  • Want to include some definitions? Minimal in what sense? Atomic in what sense? (Z,O,0) could be considered an atom for any operator $O: Z \to Z$, particularly natural ones..? So..? – Snared Apr 16 '24 at 16:36
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    minimal in the sense that it has no proper elementary substructures and atomic in the sense that for any element in the model(or tuple) the type of the element(in the theory of the model) is isolated. – Eitan Apr 16 '24 at 16:45

1 Answers1

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The fact that $(\mathbb{Z},0,+)$ is not atomic follows from the quantifier elimination for abelian groups.

The complete theory of $\mathbb{Z}$ admits quantifier elimination in the language of abelian groups, expanded by symbols $(P_n)_{n\geq 2}$, where $P_n$ picks out the multiples of $n$ in $\mathbb{Z}$. Note that this is a definitional expansion, since $P_n(x)$ can be defined by $\exists y\, \underbrace{y+\dots+y}_{n\text{ times}} = x$.

It follows that the complete type of $1$, for example, is axiomatized by the formulas $\{\lnot P_n(x)\mid n\geq 2\}$. Since no finite subset of these are sufficient to entail the others, $\mathrm{tp}(1/\varnothing)$ is not isolated.

For more details, see the answers to this question, where an explicit model $M\models\mathrm{Th}(\mathbb{Z})$ is constructed which omits $\mathrm{tp}(1/\varnothing)$ (and hence there is no elementary embedding $\mathbb{Z}\to M$, so $\mathbb{Z}$ is not a prime model for its complete theory). The model is the subgroup of $\widehat{Z}$ (the profinite completion of $\mathbb{Z}$) consisting of those elements which are divisible by all but finitely many primes.

Alex Kruckman
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  • Thank you, that's very helpful! In the answer you provided, you mentioned that the main trick involves a diagonalization procedure. However, I'm having trouble understanding how to perform diagonalization on the conjunction. Could you please elaborate or provide an example? Your assistance would be greatly appreciated. – Eitan Apr 18 '24 at 06:47
  • @EitanOri To be clear, this is diagonalization of matrices like in linear algebra, not diagonalization like in logic. I think you look at the book by Prest that I recommended if you want to see the proof. It's too long for a comment. – Alex Kruckman Apr 18 '24 at 12:15
  • I could not find the relevant part in Model Theory and Modules, you said it's in Section 2 but there are only descriptions of Ziegler spectra of various types of ring in that section. – Eitan Apr 18 '24 at 12:46
  • @EitanOri Are you looking at the 1988 book from Cambridge University Press? "Section 2.$\mathbb{Z}$" starts on page 44. – Alex Kruckman Apr 18 '24 at 13:41
  • Thank you very much I was looking at the wrong file – Eitan Apr 18 '24 at 15:50