It is well-known that the ultraproduct $\prod^\mathcal{U}_p \overline{\mathbb F_p}$, where $p$ ranges over all prime numbers and $\mathcal U$ is a non-principal ultrafilter on the set of prime numbers, is isomorphic to the field of complex numbers, because it is an algebraically closed field of characteristic $0$ and has transcendence degree continuum over $\mathbb Q$. Here $\overline{\mathbb F_p}$ denotes the algebraic closure of the field with $p$ elements.
But what happens if we bound the characteristic? More precisely, let $p$ be a prime number, let $\Lambda$ be a set and let $\mathcal U$ be a non-principal ultrafilter on $\Lambda$.
Question: What is the structure of the field $\overline{\mathbb F_p}^* = \prod^\mathcal{U}_\lambda \overline{\mathbb F_p}$?
I think, by general theorems on ultraproducts $\overline{\mathbb F_p}^*$ has to be uncountable. In particular, it must contain elements that are transcendental over $\mathbb F_p$.
I guess there is not such a smooth description as in the unbounded characteristic case, but I would appreciate any help.
Thank you in advance!
Best Answer
Well, if there characteristics are bounded, then there are only finitely many options, then there is an index set in $\cal U$ on which the characteristics is fixed, say $p$, as you note.
Since the theory of algebraically closed fields is categorical in any uncountable cardinal up to selecting the characteristics, that means that if $F$ is the ultrapower and it has size $\kappa$, then it is isomorphic to the algebraic closure of $\Bbb F_p(X)$, where $X$ is a set of size $\kappa$.
It's as simple as that. Note that this is also the case if $p=0$, once you go beyond the cardinality of the continuum, there's not much you can say.
One interesting exception, though, is when large cardinals come into play. Recall that $\kappa$ is a measurable cardinal if there is a $\kappa$-complete non-principal ultrafilter on $\kappa$. This means that not only $p$ must be fixed, but in fact on a large set, all functions are constant. So actually the ultrapower is isomorphic to the original structure. This is why the notion of a regular ultrafilter plays an important role, since it implies these sort of things don't happen.
(The issue with the functions being constant is an artifact of the closure. It is quite possible to have ultrafilters which are countably closed, but not more than that, and then the results will depend on the cardinality of the structure we are dealing with, although in this case it is countably infinite, so it would be all the same.)