# Does Keisler-Shelah isomorphism theorem hold for infinitary logics?

In model theory, the Keisler-Shelah isomorphism theorem asserts that two models of a theory are elementary equivalent if and only if they have isomorphic ultrapowers. On the other hand, assuming that $\kappa$ is strongly compact, we can ensure the existence of a $\kappa$-complete ultrafilter on every set (that is, an ultrafilter closed under intersections of less than $\kappa$ elements) and prove a version of Łoś theorem for the infinitary logic $\mathcal{L}_{\kappa, \kappa}$.

My question is whether a suitable version of the Keisler- Shelah theorem also holds for infinitary languages. Shelah's original proof (the first without using GCH), or the similar version in Chang-Keisler book "Model theory", seem to be readily generalizable, but I would like if possible to find a reference for it. The statement I have in mind is as follows:

Let $\kappa$ be a strongly compact cardinal and assume $\mathcal{A}$ and $\mathcal{B}$ are models of a theory in $\mathcal{L}_{\kappa, \kappa}$. Then $\mathcal{A}$ and $\mathcal{B}$ are elementary equivalent (with respect to the language $\mathcal{L}_{\kappa, \kappa}$) if and only there is a set $I$ and a $\kappa$-complete ultrafilter $U$ over $I$ such that the ultrapowers $\Pi_{i \in I}\mathcal{A}/U$ and $\Pi_{i \in I}\mathcal{B}/U$ are isomorphic.

Does this hold or has it been considered in the literature?

The answer is no, it doesn't necessarily hold. Let me describe a counterexample. Consider the language of linear orders. By the pigeon-hole principle, since there are only a set of possible $L_{\kappa,\kappa}$-theories in this language, there must be two ordinals $\alpha<\beta$ such that as linear orders, $\langle \alpha,<\rangle$ and $\langle\beta,<\rangle$ are $L_{\kappa,\kappa}$-elementarily equivalent.
Now consider any $\kappa$-complete ultrafilter $U$ on a set $I$. Let $j:V\to M$ be the ultrapower by $U$. The target model $M$ is well-founded, since $U$ is $\kappa$-complete. So we may take $M$ to be a transitive class.
The ultrapower $\langle\alpha,<\rangle^I/U$ is really just the same as $\langle j(\alpha),<\rangle$, and similarly $\langle\beta,<\rangle^I/U$ is isomorphic to $\langle j(\beta),<\rangle$. Since $\alpha\neq\beta$, it follows that $j(\alpha)\neq j(\beta)$, and since these are ordinals, they are not isomorphic.
So we've found two linear orders that are $L_{\kappa,\kappa}$-elementarily equivalent, but cannot have isomorphic ultrapowers by any $\kappa$-complete ultrafilter on any set.
• If $j:V\to M$ is any embedding with critical point $\kappa$, then $\langle \kappa,<\rangle$ will be an $L_{\kappa,\kappa}$-elementary substructure of $\langle j(\kappa),<\rangle$. But being distinct ordinals, they cannot have an isomorphic $\kappa$-complete ultrapower by the argument I gave. This way of arguing avoids the need for the pigeon-hole argument. – Joel David Hamkins Feb 5 '17 at 17:22