Is there a model of $Th(\frac{\mathbb{R}}{\mathbb{Z}})$ which is a periodic group?

Question

Let $L := (+,-,0)$ be the language of abelian groups and let $T = Th(\frac{\mathbb{R}}{\mathbb{Z}})$. Is there a model of $T$ which is periodic, i.e. every member of the domain has finite order. In model theoretic terms this can be restated as: Is there a model of $T$ which omits the partial type $P(x) = \{nx\neq 0 | n \in \mathbb{N}_0\} \cup T$? Since $T$ is complete this is iff $P$ is non-principal. I'm not really sure how I'm supposed to show that $P$ is non-principal in this case ( or in general). Any hints/ideas? Many thanks.

Answer 1

Both your approach and the approach in the comments work: Your type $P(x)$, expressing that $x$ has infinite order, is not isolated. Also, $\mathbb{Q}/\mathbb{Z}$ is an elementary substructure of $\mathbb{R}/\mathbb{Z}$.

To start, here's an observation that's useful whenever you want to prove that a type is not isolated (or a theory is not finitely axiomatizable).

Claim: Let $p$ be a (partial) type. If $p$ is equivalent to a single formula (relative to a theory $T$), then there are finitely many formulas $\varphi_1,\dots,\varphi_n\in p$ such that $p$ is equivalent to $\bigwedge_{i=1}^n \varphi_i$ (relative to $T$).

Proof: If $p$ is equivalent to the formula $\theta$, then $T\cup \{\theta\}\models p$ and $T\cup p\models \theta$. From the second fact, $T\cup p\cup \{\lnot \theta\}$ is inconsistent. By compactness, $T\cup \{\varphi_1,\dots,\varphi_n\}\cup \{\lnot\theta\}$ is inconsistent, for some $\varphi_1,\dots,\varphi_n\in p$. Let $\theta' = \bigwedge_{i=1}^n \varphi_i$. I claim that $p$ is equivalent to $\theta'$ (relative to $T$).

Indeed, by the inconsistency, $T\cup \{\theta'\}\models \theta$, so $T\cup \{\theta'\}\models p$. And conversely, since $\varphi_i\in p$ for all $i$, $T\cup p\models \theta'$. $\square$

Now if your type $P(x)$ is isolated, then it is isolated by $\bigwedge_{i=1}^k n_ix\neq 0$ for some $n_1,\dots,n_k>0$, so letting $N = \prod_{i=1}^k n_i$, it is isolated by $Nx\neq 0$. But this is impossible, since e.g. $\frac{1}{N+1}$ satisfies $Nx \neq 0$ but does not satisfy $P(x)$. Thus $P(x)$ is not isolated and can be omitted in some model of $T$.

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Now let's show that $\mathbb{Q}/\mathbb{Z}\preceq \mathbb{R}/\mathbb{Z}$. User mihaild's suggestion in the comments to use an Ehrenfeucht-Fraïssé game is a reasonable way to go, but let's take a different approach.

It's a theorem of Wanda Szmielew that for any abelian group $A$, the complete theory of $(A;+,-,0,(n{\mid})_{n\in \mathbb{N}})$ has quantifier elimination, where each symbol $n{\mid}$ is a unary relation symbol which holds of the elements of $A$ divisible by $n$. Note that this language is a definitional expansion of the pure group language: $n{\mid}x$ is definable by $\exists y\,(\underbrace{y+\dots+y}_{n\text{ times}} = x)$.

See the answers to this question for references and lots of other information about theories of abelian groups.

Now $\mathbb{R}/\mathbb{Z}$ is a divisible abelian group, so for every $n>0$, every element is divisible by $n$. Thus the symbols $(n{\mid})$ are interpreted trivially in $\mathbb{R}/\mathbb{Z}$, and the theory of this group actually admits QE in the abelian group language.

At this point, you could apply the Tarski-Vaught test to show $\mathbb{Q}/\mathbb{Z}\preceq \mathbb{R}/\mathbb{Z}$. This amounts to showing that for every quantifier-free formula in the language of abelian groups, $\varphi(x,b_1,\dots,b_n)$, with $b_1,\dots,b_n\in \mathbb{Q}/\mathbb{Z}$, if the formula has a realization in $\mathbb{R}/\mathbb{Z}$, then it has one in $\mathbb{Q}/\mathbb{Z}$. This shouldn't be too difficult.

But instead, for fun, let's bring in the big guns. Szmielew also classified the complete theories of abelian groups by four families of invariants (see Tim Campion's answer to the question I linked to above). In the case of a divisible group $A$, $nA = A$ for all $n>0$, so we have: \begin{align*} Tf(p;\mathbb{R}/\mathbb{Z}) = Tf(p;\mathbb{Q}/\mathbb{Z}) &= 0\\ D(p;\mathbb{R}/\mathbb{Z}) = D(p;\mathbb{Q}/\mathbb{Z}) &= 1\\ Exp(\mathbb{R}/\mathbb{Z}) = Exp(\mathbb{Q}/\mathbb{Z}) &= \infty\\ U(p,n;\mathbb{R}/\mathbb{Z}) = U(p,n;\mathbb{Q}/\mathbb{Z}) &= 0 \end{align*} It follows that $\mathbb{R}/\mathbb{Z}$ and $\mathbb{Q}/\mathbb{Z}$ are elementarily equivalent. From the definitions of the invariants we can extract the following axiomatization of their common complete theory: $A$ is a divisible abelian group of unbounded exponent such that for all primes $p$, $|\{x\in A\mid px = 0\}| = p$.

Now $\mathbb{Q}/\mathbb{Z}$ is a substructure of $\mathbb{R}/\mathbb{Z}$, and the common theory of these two groups admits quantifier elimination, so $\mathbb{Q}/\mathbb{Z}\preceq\mathbb{R}/\mathbb{Z}$.