On the prime in the ring of integers of infinite Galois extensions I need help to solve the following exercise:
Let $K$ be an infinite algebraic extension of $\mathbb Q$ and let $O_K$ denote the ring of algebraic integers in $K.$ If $\frak P$ is a prime ideal of $O_K$ then ${\frak P}\cap{\mathbb Z}$ is a prime ideal of ${\mathbb Z}.$ Moreover, if $\frak P$ is non-zero, then so is ${\frak P}\cap{\mathbb Z}.$ Thus
${\frak P}\cap{\mathbb Z}=p{\mathbb Z}$ for some prime $p.$
1) Show that $O_K/{\frak P}$ is a field, and is an algebraic extension of ${\mathbb F}_p$; then show that it is Galois over ${\mathbb F}_p$ with abelian Galois group.
2) Now suppose that $K/F$ is an infinite Galois extension. Let $\frak P,$ $O_K$ be as before and put ${\frak p}={\frak P}\cap O_F$, a prime ideal of $O_F.$ Show that the group $\text{Gal}(K/F)$ acts transitively on the set of prime ideals of $O_K$ above $\frak p.$
Thanks in advance.
 A: It suffices to know some basic Galois theory, a bit about finite fields, and a little bit of commutative algebra (the first half of Chapter 1 and the beginning of Chapter 5 in the book "Introduction to commutative algebra" by M. F. Atiyah and I. G. MacDonald is more than enough). The OP is very welcome to ask me in comments anything that seems unclear (if such a procedure is allowed here).
1. As ${\frak P}\ne0$, the quotient $L:=O_K/\frak P$ is integral over ${\mathbb F}_p$ because it is integral over $\mathbb Z$ and $p\in\frak P$. Being an integral domain over a field, $L$ is a field.
Denote by $F_p$ the algebraic closure of ${\mathbb F}_p$. It is well known (and can be easily shown) that a finite field of characteristic $p$ has $p^n$ elements (as it is an ${\mathbb F}_p$-linear space of finite dimension $n$), that, for any $n\in{\mathbb N}$, there is a unique subfield
$\text{GF}(p^n)\subset F_p$ with $p^n$ elements (here the letters $\text{GF}$ stand for Galois field), and that the Galois group of
$\text{GF}(p^n)/{\mathbb F}_p$ is a cyclic group of order $n$ generated by the Frobenius automorphism $\Phi:x\mapsto x^p$ (just because $\text{GF}(p^n)$ consists of all the roots of the polynomial $x^{p^n}-x$ which has no multiple roots).
Let $L,L'\subset F_p$ be subfields and let $\alpha,\alpha':L\to L'$ be isomorphisms. Since each of $L,L'$ is a union of finite fields $\text{GF}(p^n)$ and $\alpha\big(\text{GF}(p^n)\big)=\text{GF}(p^n)$, we obtain $L=L'$. The fact that $\alpha,\alpha'$ commute is verified easily at the level of finite subfields. Indeed, let $l\in L$. Then $l\in\text{GF}(p^n)$ for a suitable subfield $\text{GF}(p^n)\subset L$ and
$\alpha\big(\text{GF}(p^n)\big)=\alpha'\big(\text{GF}(p^n)\big)=\text{GF}(p^n)$. Since the group $\text{Gal}\big(\text{GF}(p^n)/{\mathbb F}_p\big)$ is abelian,
we obtain $\alpha\alpha'l=\alpha'\alpha l$.
2. We need the following 
Lemma (mixedmath, the answer to the question "infinite Algebraic extensions of Q"). Let $K'/F'$ be a finite Galois extension of fields algebraic over $\mathbb Q$, let $O_{K'}\subset K'$ and $O_{F'}\subset F'$ be the rings of all integers respectively in $K'$ and $F'$, and let
$0\ne{\frak p}'\triangleleft_pO_{F'}$ be a prime ideal. Then the Galois group
$\text{Gal}(K'/F')$ acts transitively on the set of all the prime ideals
${\frak P}'\triangleleft_pO_{K'}$ such that ${\frak P}'\cap O_{F'}={\frak p}'$.
Proof. Let ${\frak P}',{\frak Q}'\triangleleft_pO_{K'}$ be prime ideals such that ${\frak P}'\cap O_{F'}={\frak Q}'\cap O_{F'}={\frak p}'$ and
${\frak P}'\ne g{\frak Q}'$ for all $g\in\text{Gal}(K'/F')$. Since
$O_{K'}/{\frak P}'$ is a field by the answer to the first question, the ideal ${\frak P}'$ is maximal. Similarly, the ideal $g{\frak Q}'$ is maximal, implying ${\frak P}'+g{\frak Q}'=O_{K'}$. By the Chinese Remainder Theorem, there is $x\in{\frak P}'$ such that $x\equiv 1\mod g{\frak Q}'$ for all
$g\in\text{Gal}(K'/F')$. The norm $\prod_{g\in\text{Gal}(K'/F')}gx$ of $x$ lies in $F'$, hence, in ${\frak p}'$, and, therefore, in ${\frak Q}'$. It follows that $gx\in{\frak Q}'$, i.e., $x\in g^{-1}{\frak Q}'$ for some
$g\in\text{Gal}(K'/F')$. This contradicts $x\equiv 1\mod g^{-1}{\frak Q}'$
$_\blacksquare$
Now we are back to the infinite Galois extension $K/F$. Let
${\frak P},{\frak Q}\triangleleft_pO_K$ be prime ideals such that
${\frak P}\cap O_F={\frak Q}\cap O_F={\frak p}\ne0$. There are finite Galois extensions $K_n/F$ such that $K_0=F$, $\bigcup_{n\in{\mathbb N}}K_n=K$, and $K_n\subset K_{n+1}$ for all $n\in\mathbb N$. Denote
${\frak P}_n:={\frak P}\cap O_{K_n}$ and ${\frak Q}_n:={\frak Q}\cap O_{K_n}$. We need to find some $g\in\text{Gal}(K/F)$ such that ${\frak P}=g{\frak Q}$. Equivalently, for all $n\in\mathbb N$, we need to find $g_n\in\text{Gal}(K_n/F)$ such that ${\frak P}_n=g_n{\frak Q}_n$ and $g_n=g_{n+1}|_{K_n}$.
We construct $g_{n+1}$ by induction. As $K_{n+1}/F$ and $K_n/F$ are Galois extensions, there is some $h_{n+1}\in\text{Gal}(K_{n+1}/F)$ with $h_{n+1}|_{K_n}=g_n$. We apply Lemma to $K':=K_{n+1}$, $F':=K_n$,
${\frak p}':={\frak P}_n$, and the prime ideals
${\frak Q}':=h_{n+1}{\frak Q}_{n+1}$ and ${\frak P}':={\frak P}_{n+1}$. Note that ${\frak P}'\cap O_{F'}={\frak P}_{n+1}\cap O_{K_n}={\frak P}_n={\frak p}'$ and that
${\frak Q}'\cap O_{F'}=h_{n+1}{\frak Q}_{n+1}\cap O_{K_n}\supset h_{n+1}{\frak Q}_n\cap O_{K_n}=g_n{\frak Q}_n\cap O_{K_n}={\frak P}_n\cap O_{K_n}={\frak p}'$,
implying ${\frak Q}'\cap O_{F'}={\frak p}'$ because ${\frak p}'$ is a maximal ideal in $O_{F'}$. By Lemma, we get $g\in\text{Gal}(K_{n+1}/K_n)$ such that ${\frak P}_{n+1}=gh_{n+1}{\frak Q}_{n+1}$. It remains to put $g_{n+1}:=gh_{n+1}$.
