33
$\begingroup$

It was proved by Poonen that $\mathbb{Z}$ is definable in the structure $(\mathbb{Q}, +, \cdot, 0, 1)$ using $\forall \exists$ formula. Koenigsmann has shown that $\mathbb{Z}$ is in fact definable by universal formula. What is the simplest geometric interpretation of these results?

EDIT: It is important to note, as Joel says, that the first result in this direction was that of Julia Robinson in 1948 (Definability and decision problems in arithmetic). The references for the latest results are: Defining $\mathbb{Z}$ in $\mathbb{Q}$ (Koenigsmann's paper, arXiv version), and Characterizing integers among rational numbers with a universal-existential formula (Poonen's paper, arXiv version).

Thank you

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ See this question: mathoverflow.net/questions/19840/… @Guillaume: Here is a link to Poonen's paper lifted directly from above www-math.mit.edu/~poonen/papers/ae.pdf $\endgroup$
    – Zack Wolske
    Commented Jan 27, 2012 at 23:22
  • 16
    $\begingroup$ Probably it should be mentioned in the question that the first known definition of $\mathbb{Z}$ in $\mathbb{Q}$, very surprising at the time, was the 1948 result of Julia Robinson. Poonen's impressive result should be seen as a refinement of Robinson's theorem, lowering the complexity of the definition. $\endgroup$
    – Joel David Hamkins
    Commented Jan 28, 2012 at 0:42
  • 1
    $\begingroup$ Which language does this refer to? The language of ring theory? This should be added to the question, right? $\endgroup$
    – HeinrichD
    Commented Nov 10, 2016 at 11:58
  • 14
    $\begingroup$ @MikhailKatz "then express natural numbers as those that are sums thereof" -- that's not a first-order characterization. $\endgroup$
    – Todd Trimble
    Commented Nov 10, 2016 at 13:01
  • 11
    $\begingroup$ For example, the integers are not definable in the real field, since the latter has a decidable theory. $\endgroup$
    – Joel David Hamkins
    Commented Nov 10, 2016 at 13:16

0

Reset to default

You must log in to answer this question.