# Are there Countable $\omega$- models of ZFC?

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?

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.
• Elaborating on the last paragraph, note that there is also a huge difference between $\omega$-models and transitive models. For example, there are $\omega$-models which are low for hyperjump, whereas any transitive model has to compute basically every real you can imagine. Dec 1 '20 at 21:09