A quotient of a product of prime characteristic fields is isomorphic to a subfield of $\mathbf{C}$

Question

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 $L$.)

1. 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.

2. 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?

Answer 1

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 $\mathbb{C}.$