You seem to believe that it is somehow contradictory to have a set model of ZFC inside another model of ZFC. But this belief is mistaken.
As Gerald Edgar correctly points out, the Completeness Theorem of first order logic asserts that if a theory is consistent (i.e. proves no contradiction), then it has a countable model. To be sure, the proof of the Completeness Theorem is fairly constructive, for the model is built directly out of the syntactic objects (Henkin constants) in an expanded language. To re-iterate, since you have mentioned several times that you find something problematic with it:
- Completeness Theorem. (Goedel 1929) If a theory is consistent, then it has a model that is a set.
The converse is much easier, so in fact we know that a theory is consistent if and only if it has a set model. This is the answer to your question.
More generally, if a theory is consistent, then the upward Lowenheim-Skolem theorem shows that it has models of every larger cardinality, all of which are sets. In particular, since the language of set theory is countable, it follows that if ZFC is consistent, then it has models of any given (set) cardinality.
The heart of your confusion appears to be the mistaken belief that somehow there cannot be a model M of ZFC inside another structure V satisfying ZFC. Perhaps you believe that if M is a model of ZFC, then it must be closed under all the set-building operations that exist in V. For example, consider a set X inside M. For M to satisfy the Power Set axiom, perhaps you might think that M must have the full power set P(X). But this is not so. All that is required is that M have a set P, which contains as members all the subsets of X that exist in M. Thus, M's version of the power set of X may be much smaller than the power set of X as computed in V. In other words, the concept of being the power set of X is not absolute between M and V.
Different models of set theory can disagree about the power set of a set they have in common, and about many other things, such as whether a given set is countable, whether the Continuum Hypothesis holds, and so on. Some of the most interesting arguments in set theory work by analyzing and taking advantage of such absoluteness and non-absoluteness phenomenon.
Perhaps your confusion about models-in-models arises from the belief that if there is a model of ZFC inside another model of ZFC, then this would somehow mean that we've proved that ZFC is consistent. But this also is not right. Perhaps some models of ZFC have models of ZFC inside them, and others think that there is no model of ZFC. If ZFC is consistent, then these latter type of models definitely exist.
- Incompleteness Theorem. (Goedel 1931) If a (representable) theory T is consistent (and sufficiently strong to interpret basic arithmetic), then T does not prove the assertion "T is consistent". Thus, there is a model of T which believes T is inconsistent.
In particular, if ZFC is consistent, then there will be models M of ZFC that believe that there is no model of ZFC. In the case that ZFC+Con(ZFC) is consistent, then some models of ZFC will have models of ZFC inside them, and some will believe that there are no such models.
A final subtle point, which I hesitate to mention because it can be confusing even to experts, is that it is a theorem that every model M of ZFC has an object m inside it which M believes to be a first order structure in the language of set theory, and if we look at m from outside M, and view it as a structure of its own, then m is a model of full ZFC. This was recently observed by Brice Halmi, but related observations are quite classical. The proof is that if M is an ω-model, then it will have the same ZFC as we do in the meta-theory and the same proofs, and so it will think ZFC is consistent (since we do), and so it will have a model. If M is not an ω-model, then we may look at the fragments of the (nonstandard) ZFC inside M that are true in some Vα of M. Every standard fragment is true in some such set in M by the Reflection Theorem. Thus, by overspill (since M cannot see the standard cut of its ω) there must be some Vα in M that satisfies a nonstandard fragment of its ZFC. Such a model m = VαM will therefore satisfy all of the standard ZFC. But M may not look upon it as a model of ZFC, since M has nonstandard axioms which it thinks may fail in m.
Best Answer
Hilbert's Nullstellensatz is a consequence of the model completeness of algebraically closed fields.
Edit: I don't have a reference, but I can sketch the proof. Suppose you have some polynomial equations that don't have a solution over ${\mathbb C}$. Extend ${\mathbb C}$ by a formal solution, and then algebraically close to get a field $K$. The field $K$ obviously contains a solution, but by model completeness of algebraically closed fields, a first-order statement is true in an algebraically closed field only if it is true in every algebraically closed subfield that contains all its parameters. The existence of a solution to a finite set of polynomial equations is a first-order statement (whose parameters are the coefficients) and ${\mathbb C}$ is algebraically closed. QED.