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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 ?

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  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.

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    Ok for 1., as I thought. Concerning 2., I had already shown most of what you mention, except for the last part about the classification of algebraically closed fields of car $0$. Is this classification hard to prove, or is it of the same level ? – 2017-01-26
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    @Max: I think it comes down to a straightforward argument about transcendence degree. You just keep finding algebraically independent elements until you can't anymore, and then the rest of the field must be the algebraic closure of $\mathbb{Q}$ adjoin some set of variables. Now show that two such fields have the same cardinality iff they have the same transcendence degree (in fact I think the cardinality and transcendence degree will be the same). I imagine there is also a model-theoretic proof. – 2017-01-26
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    Well if $K$ is a field extension of $\mathbb{Q}$ of cardinality $\lambda > \aleph_0$, $[K : \mathbb{Q}] = \lambda$ (assuming AC, which implies that the cardinality of any two bases is the same even in infinite dimension, so this is well defined), and since $\overline{\mathbb{Q}}$ is countable, the transcendence degree and the cardinality will indeed be the same. So in the end it suffices to show that any two extensions with the same transcendence degree are isomorphic, and this should be pretty straightforward. Ok, thanks ! – 2017-01-26