Can second order ZFC have a set model

Second order ZFC cannot have a countable model. Can it have a set model (in full semantics) of size $\kappa$ for some cardinal ? What can be said about such a $\kappa$ ?

Solution 1:

A model of 2nd-order ZFC exists if and only if there is a strongly inaccessible cardinal.

As Eric Towers stated in the comments, for any cardinal $\kappa$, $\kappa$ is strongly inaccessible iff $(V_\kappa, \in_{V_\kappa})$ is a model of 2nd-order ZFC. So clearly, the existence of a strongly inaccessible cardinal provides a model of 2nd-order ZFC.

Conversely, it's easy to show that any model of 2nd-order ZFC must have that $\in$ is an extensional, well-founded relation. So any model of 2nd-order ZFC must be isomorphic to a transitive model of 2nd-order ZFC by Mostowski's principle. If we let $\kappa$ be the smallest cardinal not in the transitive model, then we can show the transitive model is $V_\kappa$ and that $\kappa$ is strongly inaccessible.

For an outline of that side of the proof, let $A$ be a transitive model of 2nd-order ZFC. Then let us note that the powerset function is absolute for $A$. Therefore, for all ordinals $\alpha \in A$, we see that $V_\alpha$ is absolute. Since $A = \bigcup\limits_{\alpha \in Ord(A)} V_\alpha$, we see that $A = V_\kappa$ where $\kappa$ is the smallest ordinal not in $A$ (which is clearly a limit ordinal).

Now suppose that $|\kappa| = |\alpha|$, where $\alpha$ is an ordinal in $A$. Let $R$ be a well-ordering on $\alpha$ which gives it the same order-type of $\kappa$. Then since $A$ is a model of 2nd-order ZFC, $R \in A$. Then we see that $(R, A)$ has ordering type $\kappa$, and therefore $\kappa \in A$. Contradiction. Therefore, $\kappa$ is indeed a cardinal.