By the $\text{L-S}_\downarrow$ theorem we know that there must exist a countable model of ZFC. Suppose that there is an $\omega$ model of ZFC, then would $\text{ L-S}_\downarrow$ theorem entail that there must exist an $\omega$ model of ZFC that is countable? Or all countable models of ZFC must have non standard naturals?
Are there Countable $\omega$- models of ZFC
first-order-logiclogicmodel-theorynonstandard-modelsset-theory
Related Solutions
The contradiction is inside the definition of "countable". A set is countable if there exists a surjection from $\mathbb{N}$ to our set. The function that would make our inside-the-model set countable doesn't exist inside of the model, so inside of the model, the set is uncountable.
This is a surprisingly subtle question!
Right off the bat, we have an obvious observation: if $\mathcal{M}$ is a "potential $\omega$" (= is isomorphic to the $\omega$ of some model of $\mathsf{ZFC}$), then $\mathcal{M}$ must satisfy all the arithmetic consequences of $\mathsf{ZFC}$. This set of sentences, call it "$\mathsf{ZFC_{arith}}$," is not very well understood but we can prove some basic things about it; see e.g. this old answer of mine.
That said, it's also worth noting that under a mild assumption we can whip up a formula that $\mathsf{ZFC}$ proves defines a model of $\mathsf{PA}$ which has no additional restrictions - see here - but this is quite different from the usual way arithmetic is implemented in $\mathsf{ZFC}$, which you are asking about.
But is that enough? That is, if $\mathcal{M}\models\mathsf{ZFC_{arith}}$, is $\mathcal{M}$ a potential $\omega$?
To see the issue, consider the following very artificial example. Let's say we replace the interesting arithmetic structure of the naturals with just the single equivalence relation "has the same parity as." Basically, this turns the naturals into two infinite blobs. Now consider a structure $\mathcal{A}$ in this language consisting of two infinite blobs of different cardinalities. Since the theory of two infinite blobs is complete, $\mathcal{A}$ satisfies everything $\mathsf{ZFC}$ proves about "the naturals as a pair of blobs." However, $\mathcal{A}$ is not isomorphic to any model of $\mathsf{ZFC}$'s "blobby naturals." This is because the $\mathsf{ZFC}$-theorem "the even and odd naturals have the same cardinality" does have consequences for the isomorphism type of "potential blobby naturals" but is not captured at the first-order level.
In fact, $\mathsf{ZFC_{arith}}$ falls well short of "potential $\omega$-ness" in a completely unfixable way. Barwise/Schlipf proved the following:
For a countable nonstandard $\mathcal{M}\models\mathsf{PA}$, the following are equivalent:
$\mathcal{M}$ is some $\mathsf{ZFC}$-models $\omega$.
$\mathcal{M}$ is recursively saturated.
(Strictly speaking, they proved that any structure appearing in some non-$\omega$-model of $\mathsf{ZF}$ is recursively saturated. This gives the top-to-bottom direction, with the bottom-to-top direction being folklore if I understand the history correctly. Note that countability is actually only needed in the bottom-to-top direction!)
The omitting types theorem then prevents any first-order theory from capturing potential $\omega$-ness. (Thanks to Andreas Blass for pointing this out to me in the comments below!) Depending on whether you like recursive saturation, this may or may not answer your question in the specific case of countable models. Unfortunately (or interestingly), uncountable models pose much more difficulties, and I believe no characterization is currently known even of the size-$\aleph_1$ models of $\mathsf{PA}$ which are isomorphic to some $\mathsf{ZFC}$-models' version of the naturals. (But that shouldn't be too surprising, since even the ordertypes of size-$\aleph_1$ models of $\mathsf{PA}$ aren't classified yet.)
As a coda, I vaguely recall an old paper (by H. Friedman?) on the following problem: given a first-order theory $T$, when does a tuple of formulas $\Phi$ have the property that, if we let $T_\Phi$ be the set of sentences $T$ proves holds of the structure interpreted by $\Phi$, then for all $\mathcal{A}\models T_\Phi$ there is some $\mathcal{B}\models T$ with $\mathcal{A}\cong \Phi^\mathcal{B}$? Basically, these $\Phi$s should not be "missing" any structural data (e.g. the "blobby naturals" example above was missing the structural data of the $n\mapsto n+1$ bijection). I can't seem to find the paper at the moment, however. My recent mathoverflow questions 1, 2 address this general theme, current absence of a reference notwithstanding.
Best Answer
Well, by the LS theorem, we know that if there is any model of ZFC then there is a countable model. In fact, it says there is a countable elementary submodel of the original model. By the same token, if there is a well-founded model of ZFC then LS implies it has a countable elementary submodel, which is necessarily well-founded since it is a submodel of a well-founded model. And well-founded models are $\omega$-models: the Mostowski collapse gives the relevant isomorphism.
Note however that the existence of a well-founded model is a stronger assumption than the existence of a model (but it's not that much stronger an assumption, nowhere near the strength of inaccessible cardinals). Actually, the usual argument for this ties to your question. The minimal transitive model of ZFC is $L_\alpha$ for the smallest $\alpha$ that is the height of a transitive model. In light of the discussion above, it's easy to see that it is a countable $\omega$-model of ZFC. According to this model, there are no transitive / well-founded models of ZFC for reasons you might suspect, however since it is an $\omega$-model, it agrees with the universe about arithmetic, and hence that Con(ZFC) holds, and hence that there are models of ZFC.