# Ultraproduct of pairwise non-equivalent structures

It is easy to see that the ultraproduct of a family of structures over a principal ultrafilter is elementarily equivalent to the structure whose index generates the ultrafilter, i.e. if $$U$$ is generated by $$k \in I$$, then $$\prod_{i \in I} M_i / U \equiv M_k$$. My question is, does the inverse also hold? That is, if $$\prod_{i \in I} M_i / U \equiv M_k$$ does it follow that $$U$$ is a non-principal ultrafilter that focuses on k?

Well, in general the answer is no, because we can simply consider the ultrapower of any structure over a non-principal ultrafilter. Clearly that's gonna be elementarily equivalent to that structure but the ultrafilter can still be non-principal.

Does the same hold if we consider a family of pairwise elementarily inequivalent structures? So, if $$M_i, i \in I$$ is an infinite family of structures such that $$M_i \not\equiv M_j$$ for $$i\neq j$$, does there exist a non-principal ultrafilter $$U$$ such that $$\prod_{i \in I} M_i / U \equiv M_k$$ for some $$k$$, or does $$U$$ have to necessarily be principal for this to hold?

We can perform the following sort of silly trick. If $$M_i$$ is a family of pairwise inequivalent structures ranging over an index set $$I$$ and $$U$$ is a non-principal ultrafilter on $$I$$, write $$M_{\infty} = \prod_{i \in I} M_i/U$$ for the ultraproduct. There are two cases:

Case 1: $$M_{\infty}$$ is elementary equivalent to some $$M_i$$. In this case we are done.

Case 2: $$M_{\infty}$$ is not elementary equivalent to any of the $$M_i$$. Then we can extend $$M_i$$ to a family of structures ranging over the index set $$I_{\infty} = I \cup \{ \infty \}$$ which is still pairwise inequivalent. $$U$$ induces a unique non-principal ultrafilter $$U_{\infty}$$ on $$I_{\infty}$$ given by taking all the subsets in $$U$$ and adding to them all the subsets in $$U$$ together with $$\infty$$. Now consider the ultraproduct

$$M_{\infty + 1} = \prod_{i \in I_{\infty}} M_i/U_{\infty}.$$

By Łoś's theorem, $$M_{\infty+1} \models \phi$$ iff $$X_{\infty} = \{ i \in I_{\infty} : M_i \models \phi \} \in U_{\infty}$$. By construction, this is true iff $$X = \{ i \in I : M_i \models \phi \} \in U$$, which is true iff $$M_{\infty} \models \phi$$ by a second application of Łoś's theorem. So $$M_{\infty+1}$$ is elementary equivalent to $$M_{\infty}$$.

• Thanks for the answer! This trick is actually quite cool, a lot neater than I was expecting an answer to be. I'm not sure about one direction in the elementary equivalence though. We know that $\prod_{i \in I \cup \{\infty \}} M_i / U_{\infty} \models \phi$ iff $X_{\infty} = \{ i \in I \cup \{\infty\} : M_i \models \phi \} \in U_{\infty}$. From this we have that $X = \{ i \in I : M_i \models \phi \} \in U_{\infty}$. To obtain an elementary equivalence this ought to be in $U$. Does this necessarily hold? Couldn't the extended non-principal ultrafilter we took have introduced this "big" subset? Nov 27, 2020 at 21:06
• @Ioannis: okay, I fixed it. I was being silly about the ultrafilter extension; there's a unique extension (it's the functor part of the ultrafilter functor) and the unique one makes everything work. Nov 27, 2020 at 21:37
• Yeah, I think that does it. I probably thought about it too much. Thanks again! Nov 27, 2020 at 22:52
• @Ioannis: no problem! I am too lazy to check this but I wonder if it's even the case that $M_{\infty+1}$ is isomorphic to $M_{\infty}$, not just elementary equivalent. Nov 27, 2020 at 23:01
• @QiaochuYuan: It is isomorphic, via $[(a_i)_{i\in I\cup \{\infty\}}]\mapsto [(a_i)_{i\in I}]$. Generally, changing something on a null set (or even adding more null sets) yields a (canonically) isomorphic ultraproduct. Nov 29, 2020 at 10:03