A quotient of a product of prime characteristic fields is isomorphic to a subfield of $\mathbf{C}$ This is an exercise that Serre's paper How to use Finite Fields for Problems Concerning Infinite Fields.

Let $L$ be an infinite set of prime numbers. For every $p \in L$ let $k(p)$ denote a denumerable field of characteristic $p$. Let $A = \prod k(p)$ be the product of the $k(p)$'s. Show that there exists a quotient of $A$ which is isomorphic to a subfield of $\mathbf{C}$. (Hint: Use an ultrafilter on $L$.)



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*How do you prove this? I've just started learning about ultrafilters and ultraproducts, so I don't see yet why the hint provided is a good hint. Maybe I've yet to make some realization about ultrafilters, or yet to see the right important theorem to see how to prove this. 

*Is this exercise just part of the model-theoretic Lefschetz Principal? The part that says "a first-order statement in the language or rings is true in all but finitely many algebraically closed fields of characteristic $p$ if it's true in $\mathbf{C}$" part?
 A: Consider on $L$ the filter $\mathcal{F}$ consisting of all subsets of $L$ such that the complement is finite. This filter is contained in a ultrafilter $\mathcal{U}.$ For $$a = (a_p) \in \Pi_{p \in L} k(p)$$ let $$Z(a) = \{p \in L | a_p = 0 \}.$$ We let $M_\mathcal{U} \subset A$ consist of those $a \in A$ such that $Z(a) \subset \mathcal{U}.$ This gives a maximal ideal on $A,$ and thus, $K = A /M_\mathcal{U}$ is a field. Note that $\oplus_{p \in L } k(p)\subset M_\mathcal{U}.$
Claim: K has characteristic zero.
Suppose the characteristic is $q>0.$ Then we would have that the element $q=(q,q, \ldots) \in A$ would be in $M_\mathcal{U},$ but this can not happen. Indeed, if the characteristic $q \neq p,$ for $p \in L,$ then $q= (q,q, \ldots ,) \in A $ has no zero component, so $Z(q) = \emptyset,$ and $\emptyset \not \in \mathcal{U}.$ If $q = p$ for some $p \in L,$ then $q = (q,q , \ldots)$ satisfies $Z(q) = \{p\},$and $\{p\} \not \in \mathcal{U}.$ Thus, $K$ has characteristic zero.  
Now, it is easy to see that the cardinality of $K$ is at most uncountable. We will then be done if we note the following.
Claim: Any field $K$ of characteristic zero of cardinality at most the continuum can be embedded into $\mathbb{C}.$
Indeed, this is obvious if $K$ is not transcendental over $\mathbb{Q}.$ Suppose now that $K$ is transcendental over $\mathbb{Q}$ and let $S_i \in K,$ $i \in S$ be a transcendence basis. Clearly the cardinality of $S$ is at most the continuum. If we now pick a transcendence basis $T_j,j \in T $ of $\mathbb{C}$ over $\mathbb{Q},$ one can find an injection $S \rightarrow T.$ This gives us a field embedding of $\mathbb{Q}((S_i)_{i \in S}) \rightarrow \mathbb{Q}((T_j)_{j \in T}).$ The latter field is a subfield of $\mathbb{C}.$ Now extend this embedding by adjoining the algebraic elements to $K$ and get an embedding into $\mathbb{C}.$ 
