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?


1 Answer 1


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 $M_{\infty}$.

  • $\begingroup$ 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? $\endgroup$ Nov 27, 2020 at 21:06
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    $\begingroup$ @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. $\endgroup$ Nov 27, 2020 at 21:37
  • $\begingroup$ Yeah, I think that does it. I probably thought about it too much. Thanks again! $\endgroup$ Nov 27, 2020 at 22:52
  • $\begingroup$ @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. $\endgroup$ Nov 27, 2020 at 23:01
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    $\begingroup$ @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. $\endgroup$
    – tomasz
    Nov 29, 2020 at 10:03

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