# Isomorphic ultraproducts of models implies isomorphic models?

Suppose we have two countable models $\mathcal{M},\ \mathcal{N}$ of a theory $T$,in a countable languagle $\mathcal{L}$. We take an ultrafilter $U$ of $\aleph_0$ and find that the ultraproducts

$$\mathcal{M}^{\aleph_0}/U \cong \mathcal{N}^{\aleph_0}/U$$

are isomorphic. One can show that if $U$ is countably complete, then $\mathcal{M}\cong \mathcal{N}$. Must this isomorphism exist even if $U$ is not countably complete?

No. In fact, the celebrated Keisler-Shelah theorem tells us that whenever $\mathcal{M}$ and $\mathcal{N}$ are elementarily equivalent countable structures, there is an ultrafilter $U$ on $\aleph_0$ such that $\mathcal{M}^{\aleph_0}/U\cong \mathcal{N}^{\aleph_0}/U$.
By the way, you mention the case when $U$ is countably complete. But the only countably complete ultrafilters on $\aleph_0$ are the principal ultrafilters. If you want a countably complete nonprincipal ultrafilter, you need to look at ultrafilters on $\kappa$, where $\kappa$ is a measurable cardinal. And even then the implication "isomorphic ultrapowers implies isomorphic" is somewhat trivial, since if $\mathcal{M}$ is countable and $U$ is countably complete, then $\mathcal{M}^I/U\cong \mathcal{M}$, and we have $\mathcal{M}\cong \mathcal{M}^I/U \cong \mathcal{N}^I/U \cong \mathcal{N}$.