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?
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Any theory that is uncountably categorical would do the trick. – Asaf Karagila Jun 10 '19 at 20:39
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@AsafKaragila But will any free ultrafilter necessarily give us the same cardinality for the quotient? – George Peterzil Jun 10 '19 at 20:41
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It's not true in general that ultrapowers do not depend on the ultrafilter. – Eric Wofsey Jun 10 '19 at 20:41
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@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? – George Peterzil Jun 10 '19 at 20:43
1 Answers
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.
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Very cool! Thank you. So, the question of which finite extensions of the rationals can be realized as such an ultraproduct is simply a question of whether or not the polynomial equations defining them occur in compatible indices of primes infinitely many times? – George Peterzil Jun 10 '19 at 20:50
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@George See also my convoluted proof that $\sqrt2$ is irrational. – Asaf Karagila Jun 10 '19 at 20:57
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You mean which finite extensions can be realized as subfields of such an ultraproduct? Then yes, that's right. (Or taking a primitive element, you only need to look at one equation.) – Eric Wofsey Jun 10 '19 at 21:01