This was a problem in an exam I took last semester, but I never got the chance to ask my professor how to solve it. Here goes:
Let $\kappa$ be an inaccessible cardinal. Then $V_\kappa$ is a transitive model of ZFC (with the usual $\in$). Let $U$ be a non-principal ultrafilter over $\kappa$ and let $\tilde{M} = (V_\kappa)^\kappa/U$ be the corresponding ultrapower of $V_\kappa$.
We say that an ultrafilter $U$ has the $\sigma$-property if given sets $A_i \subset \kappa, i \in \mathbb{N}$, we have that if $A_i \in U$ for every $i \in \mathbb{N}$ then $\bigcap_{i \in \mathbb{N}} A_i \in U$.
(a) Show that $\tilde{M}$ is well-founded iff $U$ has the $\sigma$-property.
(b) Suppose $\tilde{M}$ is well-founded, and let $M$ be obtained from $\tilde{M}$ through the Mostowski collapse (so that $M$ is a transitive model of ZFC). Show that $M$ is elementarily equivalent to $V_\kappa$ but $M \neq V_\kappa$.
IDEAS: One of the definitions of well-founded (and I believe the best one for this problem) is: A model of ZFC $(M, \varepsilon)$ is NOT well-founded iff there exists a sequence $x_1, x_2, \ldots$ in $M$ (note that we're talking about a ''true'' sequence and not a sequence of the model) with $x_{n+1}$ $\varepsilon$ $x_n$ for all $n \in \mathbb{N}$.
I managed to prove the easier direction in part (a): Suppose $\tilde{M}$ is not well-founded. Then there exists a sequence such that $\ldots$ $\varepsilon$ $x_2$ $\varepsilon$ $x_1$. Let $f_i$ be an element in the equivalence class $x_i$. By definition, $x_{i+1}$ $\varepsilon$ $x_{i} \Leftrightarrow A_i := \{k \in V_\kappa | f_{i+1}(k) \in f_i(k) \} \in U$.
Since $V_\kappa$ is well-founded, $\bigcap_{i \in \mathbb{N}} A_i = \emptyset \notin U,$ thus $U$ does not have the $\sigma$-property.
Also, in part (b), $M$ is clearly elementary equivalent to $V_\kappa$ since it is isomorphic to $\tilde{M}$ (through the Mostowski collapse), which is elementary equivalent to $V_\kappa$.
Any thoughts on the other parts?