TITLE: The residual field of $\operatorname{Spec}(\prod_{p} \overline{\mathbb{F}_p})$.
QUESTION [5 upvotes]: Let $I$ be a set and $\mathfrak{U}$ be the set of all ultrafilters on $I$. If $E \subset I$ we define $\mathfrak{U}_E$ to be the set consisting of those elements of $\mathfrak{U}$ which contain $E$. We can endow $\mathfrak{U}$ with a topology for which a basis of open sets is given by the $\mathfrak{U}_E$'s.
Let $(k_i)_{i\in I}$ be family of fields and put $A = \prod_{i \in I} k_i$. We can then show that for each $\mathscr{F} \in \mathfrak{U}$ $ \mathfrak{m}_{\mathscr{F}} = \{ (x_i)_{i\in I} \in A \;|\; \{i \in I \;|\; x_i = 0 \} \in \mathscr{F} \}$ is a maximal ideal of $A$. 
Furthermore the map $\mathscr{F} \mapsto \mathfrak{m}_{\mathscr{F}}$ is a homeomorphism from $\mathfrak{U}$ to \operatorname{Spec}(A)$.
I've more or less been able to show all of this. But now I'm asked (this is from an old exam from my algebraic geometry course) to show that if $I$ is the set of prime numbers, $k_i = \overline{\mathbb{F}_i}$ and $\xi \in \operatorname{Spec}(A)$ is a point corresponding to a non principal ultrafiler of $I$ then $\kappa(\xi)$ (the residual field of $\xi$) is an algebraically closed field of characteristic zero.
I haven't been able to prove this so I'm wondering if anyone can suggest a solution. I'm in fact even more interested in a reference for this result because I find it really cool (and I could check wether what I've done is correct).
For information this is exercise $4$ of this exam (in french), you will find more details about ultrafilters in there : http://www.math.jussieu.fr/~lepage/TD/GeoAlg/termgeo-alg2012.pdf

REPLY [3 votes]: That is in fact a very simple exercise in model theory.
First one has to write down the first-order theory of fields, and add the schema specifying that a field is algebraically closed. That is, $\forall x_0\ldots\forall x_n\exists y(x_ny^n+\ldots+x_1y+x_0=0)$.
Now Łoś theorem tells us that the ultraproduct is a field, and that it is in fact algebraically closed. This is because the ultraproduct satisfies a sentence if and only if the set of indices of structures satisfying this sentence is large (in the ultrafilter). And all of them satisfy these part of the theory.
To see that the characteristics is $0$, note that for every $p$ the set of fields whose characteristic is larger than $p$ is large (in fact cofinite). Therefore the product satisfies $1+1\neq 0$ and $1+1+1\neq 0$ and so on. Which is the same thing as saying that it has characteristic $0$.
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