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!
 A: 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.]
