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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?

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    $\begingroup$ I posted an answer. To see that the quotient is a field, use that an ultrafilter gives a maximal ideal. Of course, you already know that this is what you need to show. Prove that there is a bijection between filters of $L$ and ideals of $A.$ Then the claim that an ultrafilter is a maximal ideal is trivial. $\endgroup$
    – Dedalus
    May 20, 2018 at 16:59
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    $\begingroup$ The quotient is countable if you quotient out by a principal ultrafilter (since then you will get $k(p)$!) but otherwise I believe it is always uncoutnable. But this is not important for what you want to show, you only need that the cardinality is at most uncoutnable. $\endgroup$
    – Dedalus
    May 20, 2018 at 17:00
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    $\begingroup$ @Dedalus Oh, thinking of filters as ideals helps a lot. Don't know why I didn't make that connection before. :P Thank you! I'm going to work through the details of your answer to make sure I'm comfortable everything, but it all looks reasonable. While I'm doing that, you wouldn't happen to have any insight on the second part of my question would you? $\endgroup$
    – Mike Pierce
    May 20, 2018 at 17:19
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    $\begingroup$ I must admit that I am not sure. What is the first-order statement you have in mind? You can use what is called Łoś's theorem en.wikipedia.org/wiki/Ultraproduct#%C5%81o%C5%9B's_theorem to prove that $K$ is of characteristic zero (since the equation $px-1$ has solutions in all $k(p)$ for all but finitely many $p$). If you would further suppose that each $k(p)$ is algebraically closed and countable in your question, then you could prove that $\Pi_{p \in L} k(p)/\mathcal{M}_U$ is actually isomorphic to $\mathbb{C}!$ $\endgroup$
    – Dedalus
    May 20, 2018 at 17:27
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    $\begingroup$ @Dedalus Yeah! After you establish the isomorphism $\Pi_{p \in L} k(p)/\mathcal{M}_U \simeq \mathbb{C}$ Łoś' theorem takes care of that part of the Leschetz principal easily. Cool! Thank you. :) $\endgroup$
    – Mike Pierce
    May 23, 2018 at 18:01

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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 $\mathbb{C}.$

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