I can't imagine how this is possible:
Let $\mathcal{M}$ be a nonstandard model of arithmetic. Show that:
- There is an element $a\in M$ such that for all prime numbers $p$, we have that $\mathcal{M} \vDash S^p0|a$.
- There is an element $a\in M$, $a>1$ such that for no prime number $p$, we have that $\mathcal{M} \vDash S^p0|a$.
Thank you so much! :)
Let $\varphi$ be the sentence that says that for every $x\gt 0$ there is a $y\gt 0$ such that $t$ divides $y$ for every $t$ in the interval $0\lt t\le x$.
This sentence is not difficult to write down in the usual language of first-order arithmetic, and is true in the natural numbers.
Now let $M$ be a non-standard model of "true arithmetic," and let $m\in M$ be larger than every standard integer. Then since $\varphi$ is true in $M$, there is an $a$ such that $t$ divides $a$ for every $t$ in the interval $0\lt t\le m$. In particular, every standard prime divides $a$. (Since $\varphi$ is a theorem of first-order Peano arithmetic, the same argument works for models of that theory.)
For the second part, instead of the $a$ of the previous argument, use $a+1$.