4
$\begingroup$

I have this exercise which seems very simple, but I am not able to find a solution.

Given a structure $\mathfrak A$=($\Bbb Q$, $\cdot$)

Is $\Bbb Q_{\ge 0}$ elementary definable in $\mathfrak A$?

The formula $\phi(x)=\exists y(x=y \cdot y)$ would not accept 2 which is obviously in $\Bbb Q_{\ge 0}$. On the other side I am not able to find an automorphism which is not compatible with the relation $\Bbb Q_{\ge 0}$.

Can you help me to find such a formula or the automorphism?

$\endgroup$
  • $\begingroup$ The question asks if it is definable. Maybe it is not... Have you tried to show that it is not definable? Do you have any information that makes you assume that it is indeed definable? $\endgroup$ – Dirk Aug 3 '17 at 9:55
3
$\begingroup$

Any nonzero rational number $q$ can be written uniquely as $q = 2^nr$, where $n$ is an integer and $r$ is a rational number, such that when $r$ is written in lowest terms as $a/b$, both $a$ and $b$ are odd. Define $v_2(q) = n$. (This is called the $2$-adic valuation of $q$.)

Consider the map $f\colon \mathbb{Q}\to\mathbb{Q}$ defined by $$f(q) = \begin{cases} q & \text{if $v_2(q)$ is even}\\ -q & \text{if $v_2(q)$ is odd}.\end{cases}$$

Now check that $f$ is an automorphism of $(\mathbb{Q},\cdot)$ which doesn't preserve $\mathbb{Q}_{\geq 0}$, since $f(2) = -2$.

Edit: I feel like I should put some model theory in this answer, so I'll point out that $(\mathbb{Q},\cdot)$ is an abelian group (well, almost - it has the extra element $0$, but this element interacts trivially with the rest of the structure and doesn't affect the definable sets at all), and definable sets in abelian groups (and $R$-modules more generally, for any ring $R$) are well-understood. See Theorem 3.3.5 in Tent & Ziegler's book A Course in Model Theory.

Every formula is equivalent to a Boolean combination of positive primitive formulas, which have the form $$\exists y_1,\dots,y_n\, \bigwedge_{i=1}^k \varphi_k(\overline{x},\overline{y}),$$ where each $\varphi_k$ is an equation. In the case of $(\mathbb{Q}_{\neq 0},\cdot)$, an equation looks like $$x_1^{a_1}\cdot \ldots \cdot x_m^{a_m}\cdot y_1^{b_1}\cdot \ldots \cdot y_n^{b_n} = 1,$$ where all the $a_i$ and $b_i$ are integers.

So if you want to check whether a particular subset of $\mathbb{Q}$ is definable, you can do this by checking whether or not its a finite Boolean combination of sets defined by positive primitive formulas. It turns out that positive primitive formulas always define subgroups, which can be helpful (though a Boolean combination of subgroups is not necessarily a subgroup). Of course, in this case the argument via an automorphism was much easier.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.