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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?

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This is a very nice question.

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.

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  • $\begingroup$ Thanks Joel. I was wondering, what if we require the stronger hypothesis that one of the models is an elementary substructure of the other? Would that be enough for having isomorphic ultrapowers or you can somehow modify your counterexample? $\endgroup$ – godelian Feb 5 '17 at 15:09
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    $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Feb 5 '17 at 17:22

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