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I'm working on some exercises in model theory, but on this one I don't know how to start. Please help to solve this.

Prove that $\text{Th}(\mathbb{Z},+)$, the theory of the structure $(\mathbb{Z},+)$, has uncountably many $1$-types.

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Do you mean complete 1-types? When are two 1-types equal? –  Colin McQuillan Jan 30 '13 at 12:28
    
I mean complete 1-types. For $\Gamma = \Gamma (x)$ a set of sentences: $\Gamma$ is a (complete) 1-type iff every finite subset of $\Gamma$ is realizable and for every formula $\phi = \phi (x)$ either $\phi$ or $\neg \phi$ belongs to $\Gamma$. –  natural Jan 30 '13 at 12:48

1 Answer 1

A hint is: what sorts of sets can you define in this theory? Think about $\phi(x)=\exists y . x+py=c$.

A direct answer is given by noticing that, for any $c\in \prod_p \mathbb Z/p\mathbb Z$ where $p$ ranges over the primes, every finite set of formulas of the form $\exists y . x+py=c_p$ is realisable. By Zorn's lemma we can pick a complete 1-type $t_c$ extending the set $\{\exists y . x+py=c_p \mid p\text{ prime}\}$, and if $c\neq c'$ then $t_c\neq t_{c'}$. If you want to remove the use of the axiom of choice, you'll have to define the complete 1-types directly by being more careful (I suggest looking up profinite integers).

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