If a first-order theory $T$ has an infinite model, does $T$ necessarily have two isomorphic models that look non-isomorphic inside a subuniverse?

Assume a proper class of inaccessibles.

I find the following general question interesting: for which isomorphism classes $C$ of first-order structures sharing a common signature does there exist a transitive model $M$ of ZFC such that $C \cap M$ has elements which look non-isomorphic from the perspective of $M$? A related question is whether every first-order theory $T$ that admits an infinite model gives rise to such isomorphism classes. Explicitly:

Question. Suppose we're given a first-order theory $T$ that admits an infinite model. Does there necessarily exists a transitive model $M$ of ZFC together with $T$-models $X,Y \in M$ such that $X$ and $Y$ are non-isomorphic in $M$, despite that they're isomorphic in the ambient universe $V$?

Solution 1:

I will use your assumption of the existence of a proper class of inaccessible just to get models of $\mathsf{ZFC}$; however, if you just require models of $\mathsf{ZFC}$ without power set, you can remove any large cardinal assumptions.

Find a countable language $\mathscr{L}$ and two $\mathscr{L}$-structures $A$ and $B$ such that $A \equiv_{\infty \omega} B$ and $A$ and $B$ have different cardinalities. Let $\Theta$ be an inaccessible cardinal such that $V_\Theta$ contains $\mathscr{L}$, $A$, and $B$. Let $N \prec V_\Theta$ be a elementary substructure containing $A$ and $B$. Since $\Theta$ is inaccessible, $N \models \mathsf{ZFC}$. Let $\pi : N \rightarrow M$ be the Mostowski collapse map. Let $A' = \pi(A)$ and $B' = \pi(B)$. $M$ thinks that $A'$ and $B'$ are $\pi(\mathscr{L})$-structures. $M$ thinks $A' \equiv_{\infty \omega} B'$. Since $V_\Theta$ thinks that $A$ and $B$ are not isomorphic, $M$ thinks $A'$ and $B'$ are not isomorphic. Since $M \models \mathsf{ZFC}$, it is an admissible set. $M \models A' \equiv_{\omega, \infty} B'$, then implies that $A' \equiv_{\infty\omega} B'$ in $V$ (by a result of Mark Nadel, I believe). In $V$, $A'$ and $B'$ are countable. So by Scott Theorem, they are isomorphic in $V$.