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?


1 Answer 1


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

So any pair of countable structures which are elementarily equivalent but not isomorphic provides a counterexample.

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 $\mathcal{M}\cong \mathcal{M}^I/U \cong \mathcal{N}^I/U \cong \mathcal{N}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.