If $M$ is any model of ZFC then by Infinity it has a set $\omega^M$. Also $\text{ZFC} \vdash (\omega \models \text{PA})$, so $M \models (\omega \models \text{PA})$. However, what can we say externally about $\omega^M$? Does it satisfy PA? Basically, what can we say externally about the natural numbers of an arbitrary model of ZFC?
Also, given a model of ZFC, can we get another model of ZFC with (externally) natural numbers isomorphic to any chosen nonstandard model of PA?
Best Answer
Yes, $\omega^M$ satisfies PA. More generally, suppose $X$ is any first-order structure internal to $M$. Then a simple induction on formulas shows that for any first-order formula $\varphi(x_1,\dots,x_n)$ and any $a_1,\dots,a_n\in X$, $M\models(X\models\varphi(a_1,\dots,a_n))$ iff $X\models\varphi(a_1,\dots,a_n)$. (To be clear, when I say $X\models\varphi(a_1,\dots,a_n)$, I really mean $X'\models\varphi(a_1,\dots,a_n)$ where $X'=\{a\in M:M\models a\in X\}$ equipped with the first-order structure defined using the internal first-order structure of $X'$.)
However, $\omega^M$ cannot be an arbitrary nonstandard model of PA. For instance, since ZFC proves the consistency of PA, $M\models(\omega\models Con(PA))$ and thus $\omega^M$ must satisfy $Con(PA)$.
(There is an important subtlety here, which is that the argument of the first paragraph only applies when $\varphi$ is an actual first-order formula, i.e. an external one in the real universe, rather than a first-order formula internal to $M$. If $\omega^M$ is nonstandard, then $M$ will have "first-order formulas" whose length is nonstandard and therefore are not actually formulas from the external perspective. This means, for instance, that if $X$ is a structure internal to $M$ which is externally a model of PA, then $M$ may not think $X$ is a model of PA, since $M$ has nonstandard axioms of PA which $X$ may not satisfy. See this neat paper for some dramatic ways that things like this can occur.)