Reduced product isomorphic to $\mathbb{C}$

Question

I've read many times that if $\mathcal{U}$ is a non principal ultrafilter over $\mathbb{N}$, and if we let $\overline{\mathbb{F}_p}$ denote the algebraic closure of $\mathbb{F}_p$, then $\prod_{p \in \mathbb{P}} \overline{\mathbb{F}_p} / \mathcal{U}$ is isomorphic to $\mathbb{C}$. I have two questions about this theorem : When we define the reduced product, $\mathcal{U}$ is usually an ultrafilter over the index set : here do we consider the ultrafilter carried from $\mathbb{N}$ to $\mathbb{P}$ through a bijection, or the set $\mathcal{F} = \{A\cap \mathbb{P} \mid A \in \mathcal{U} \}$ ? I think it would have to be the first one, as it's not sure whether $\mathbb{P} \in \mathcal{U}$, and if it doesn't belong to it, then $\emptyset \in \mathcal{F}$ (as $\mathcal{U}$ is ultra). So I'm pretty sure about the answer to this one,but I'd rather check. The second question is : how does the proof go ? Is it easy (uses basic knowledge about ultraproducts and fields) or is it a complicated proof ?

Answer 1

1. Yes, the intended meaning is to transport the ultrafilter to an ultrafilter over the primes.

2. Łoś's theorem can be used to show that this ultraproduct is an algebraically closed field of characteristic zero, and it's not hard to see that it has the same cardinality as $\mathbb{C}$. Now, it turns out that algebraically closed fields of characteristic zero are classified by their cardinality: two such fields are isomorphic iff they have the same cardinality.