# Algebraic closure of the rational inside a quotient of product of finite fields

I'm trying to solve the following exercise:

" Consider the ring $R = \prod_{p} \mathbb{F}_p$, where $p$ runs over all prime numbers and $\mathbb{F}_p$ is a field with $p$ elements. Show that there exists a maximal ideal $M$ of $R$ such that the field $R/M$ contains (a field isomorphic to) the algebraic closure $\mathbb{Q}^\text{ac}$ of the rational numbers. "

Actually, I have no idea. A hint suggests to use the Chebotarev's density theorem, but I don't see any relation with it.

Thank you very much for any suggestion/reference!

• Well, first of all, that is not the product of all finite fields... Commented Jan 4, 2015 at 14:36
• @ThomasAndrews Come on, we know that for each prime $p$ there exists exactly one field of $p$ elements, up to isomorphism. Anyway, I changed the title. Commented Jan 4, 2015 at 15:39
• But there is also a field of order $p^2$, $p^3$, etc. @Marvolo The prime fields are not the only finite fields. Commented Jan 4, 2015 at 15:50
• Commented Feb 17, 2017 at 15:05

A bit clumsy, but it does use Chebotarev's theorem :

Let $P$ be the set of all primes and for $f\in\mathbb Z[x]$ monic and irreducible let $P_f\subset P$ be the set of primes $p$ which don't divide the discriminant of $f$ and such that $f$ splits to linear factors in $\mathbb F_p[x]$. Chebotarev now says that for any finite collection $f_1,\dots,f_n$ we have $P_{f_1}\cap\dots\cap P_{f_n}\neq\emptyset$ (by taking the splitting field $K$ of $f_1f_2\dots f_n$ and considering the primes $p$ for which the corresponding Frobenius automorphism of $K$ is the identity - they have non-zero density).

Since, as shown above, $P_f$'s satisfy the finite intersection property, there is an ultrafilter $\mathcal U$ on $P$ such that $P_f\in\mathcal U$ for every $f$. The ultrafilter $\mathcal U$ is not principal (any prime divides the discriminant of some $f$, hence $\bigcap_f P_f=\emptyset$)), so the ultraproduct $\prod_p\mathbb F_p/\mathcal U$ is a field of characteristic $0$, and by construction every $f$ splits in this field. [It is a quotient of $\prod_p\mathbb F_p$, so it can be written as $(\prod_p\mathbb F_p)/M$ where $M$ is a maximal ideal.]

• Thank you very much for your answer. I have a couple of questions. Do you use Los theorem applied to the formula $\varphi_f \equiv \exists y_1 \cdots \exists y_{deg(f)} f(x) = (x-y_1)\cdots(x-y_{deg(f)})$ to prove that each $f$ splits in the ultraproduct? Is it $M=\{(a_p) : \{p \in P : a_p = 0\} \in U\}$, right? Thus $M$ is maximal since $R / M = \prod_p \mathbb{F}_p / U$ is a field. Commented Jan 5, 2015 at 11:43
• @Marvolo yes to both questions. I probably should have written the answer without ultrafilters: if $I_f=\{(a_p):a_p=0\,\forall p\notin P_f\}$ then there is by Zorn lemma a maximal ideal $M$ containing all $I_f$'s, and $f$ splits in $R/M$ since it splits in $R/I_f$. But I like ultraproducts :) Commented Jan 5, 2015 at 12:53