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?



Defining $\mathbb{Z}$ in $\mathbb{Z}^\ast/P\mathbb{Z}^\ast$ where $P$ is an infinite prime
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.


