This came up in the comments to [an answer of Joel's](https://mathoverflow.net/questions/450222/ultraproducts-in-the-category-of-structures-and-elementary-embeddings). Suppose $\mathcal{M}\_i$ ($i\in I$) are elementarily equivalent structures in the same fixed signature and $\mathcal{U}$ is an ultrafilter on $I$. Must some $\mathcal{M}\_i$ elementarily embed into $\prod\_{i\in I}\mathcal{M}\_i/\mathcal{U}$?


I'm happy to add further restrictions to make this question easier, such as setting $I=\omega$ or requiring the language of the structures to be finite.



Can ultraproducts avoid all "factor structures"?
Here is another example, which is inspired by James's answer, but I find this one a little simpler.


Let $T$ be the theory of an equivalence relation $\sim$ with infinitely many classes, all infinite, plus countably many constants $c\_0,c\_1,c\_2,\ldots$, taken from different equivalence classes. This is a complete theory, which can be seen by elimination of quantifiers.


Let $M\_n$ be a model with all equivalence classes countable, except the equivalence class of $c\_n$, which we take of size $\mathfrak{c}^+$. Let $M=\prod\_n M\_n/\mu$ be the ultraproduct of the $M\_n$ by a nonprincipal ultrafilter $\mu$ on $\omega$.


The size of the equivalence class of $c\_n$ in $M$ is determined by functions from $\omega$ to the corresponding $[c\_n]^{M\_i}$, which are all countable except for $M\_n$. Since no one factor is relevant for the nature of $M$, it follows that $[c\_n]^M$ has size continuum. Thus, $M\_n$ does not embed in $M$.


So none of the factors embed into the ultrapower, as desired.


**Optimal size.** For the reasons mentioned in [Emil's comment below](https://mathoverflow.net/questions/450241/can-ultraproducts-avoid-all-factor-structures#comment1163705_450257), if CH holds this size $\mathfrak{c}^+$ is the best possible. I am unsure whether there could be a smaller counterexample if CH fails.


**Analyzing the ultrapower further.** I find it interesting to notice a few things about the ultrapower structure — in fact we can describe it exactly. It will have a bunch of other equivalence classes not containing any of the constants $c\_n$. Any function $f$ with $f(n)$ not almost always chosen from the class of the same constant $c\_k$ will represent a new equivalence class in the ultrapower. These will all have size continuum also (and there are continuum many of them), except for the one equivalence class arising from functions $f$ with almost always $f(n)\in[c\_n]^{M\_n}$, choosing from the big class at these coordinates. This one equivalence class in the ultrapower will have size $(\mathfrak{c}^+)^\omega=\mathfrak{c}^+$.


So the ultraproduct $M$ consists of continuum many equivalence classes of size continuum, with countably many of them occupied by constants $c\_n$, plus one more equivalence class of size $\mathfrak{c}^+$.


**Generalizing to arbitrary ultrafilters.** The construction generalizes to any uniform ultrafilter $\mu$ on any set $I$. We have an equivalence relation with infinitely many classes, all infinite, plus constants $c\_i$ for each $i\in I$, and inequivalent. Let $M\_i$ be a model with all equivalence classes countable, except for $[c\_i]^{M\_i}$, which should have size larger than $\omega^I$. The ultrapower $M=\prod\_i M\_i/\mu$ will have $[c\_i]^M$ of size at most $\omega^I$, and so no $M\_i$ will embed into $M$.


