# The ultraproduct of all prime fields

In the process of reviewing a proof of the Ax-Grothendieck theorem using ultraproducts, I came across a fun little fact: given a free ultrafilter $$\mathcal U$$ on the natural numbers, one necessarily has $$\mathbb C\cong \prod_{primes}\mathbb{\bar F}_{p}/\mathcal U$$ (the product of all algebraic closures of the prime fields), regardless of the choice of filter (as long as it's not principal). Even more surprising than being isomorphic to the complex field, they are all isomorphic to each other. My question is this: other than ultrapowers, which clearly all give rise to the same structure, what other ultraproducts (I'm interested mostly in algebraic examples) give rise to the same first-order structure, regardless of the specific free ultrafilter picked? Specifically, are all products $$\prod_{primes}\mathbb F _{p}/\mathcal U$$ (with $$\mathcal U$$ free) isomorphic?

• Any theory that is uncountably categorical would do the trick. Jun 10, 2019 at 20:39
• @AsafKaragila But will any free ultrafilter necessarily give us the same cardinality for the quotient? Jun 10, 2019 at 20:41
• It's not true in general that ultrapowers do not depend on the ultrafilter. Jun 10, 2019 at 20:41
• @EricWofsey OK, yeah, that relates to my question in the comments above. Can one construct two free ultrafilters that give rise to quotients of different cardinality? Or can some other second-order properties get screwed up in the way? Jun 10, 2019 at 20:43

Ultraproducts $$\prod_{primes}\mathbb F _{p}/\mathcal U$$ very much depend on the ultrafilter. For instance, if you take any polynomial $$f$$ with integer coefficients such that $$f$$ has roots mod infinitely many primes but does not have roots mod infinitely many other primes (for instance, $$f(x)=x^2+1$$), then whether $$f$$ has a root in the ultraproduct depends on which of these sets of primes is in the ultrafilter.
More generally, given any family of structures $$(M_i)$$, if there is a sentence $$\phi$$ which is true in infinitely many of the $$M_i$$ and also false in infinitely many of the $$M_i$$, then $$\phi$$ can be either true or false in an ultraproduct of the $$M_i$$ by a nonprincipal ultrafilter, so not all such ultraproducts are even elementarily equivalent.
• @George See also my convoluted proof that $\sqrt2$ is irrational. Jun 10, 2019 at 20:57