2
$\begingroup$

In continuation of my recent questions, here is the last one: Is there a simple formula preferably existential that defines $\mathbb{Z}$ in $\mathbb{Z}^\ast/P\mathbb{Z}^\ast$ where $\mathbb{Z}^\ast$ is an elementary extension of $\mathbb{Z}$ and $P$ is an infinite prime?

$\endgroup$
3
$\begingroup$

No. If $\phi(x,y)$ is any formula with two free variables $x$ and $y$, then it's true $\mathbb Z$ and therefore also in the elementary extension $\mathbb Z^*$ that $(\forall y)$ [if $\phi(0,y)$ and if $(\forall x)\ (\phi(x,y)\implies \phi(x+1,y)\land\phi(x-1,y))$ then $(\forall x)\ \phi(x,y)$]. Now if $\psi(x)$, with only $x$ free, defined $\mathbb Z$ in $\mathbb Z^*/P\mathbb Z^*$, then the pre-image $D$ of $\mathbb Z$ in $\mathbb Z^*$ would have a definition with $P$ as a parameter, say $\phi(x,P)$, in $\mathbb Z^*$, to which we could apply the induction principle above. Since $D$ contains 0 and is closed under adding and subtracting 1, it would have to be all of $\mathbb Z^*$, which it is not.

$\endgroup$
1
  • $\begingroup$ Oops. I just realized that my answer here is essentially a repetition of answers by Joel Hamkins and Emil Jerabek to previous questions from the same OP. I'm not sure whether I should delete my answer here, or leave it up in the hope that enough repetition will make the matter clear to the OP. $\endgroup$ – Andreas Blass Feb 3 '12 at 23:55

Your Answer

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