Ramification in the ring of all algebraic integers If $F$ is a finite extension of $\mathbb{Q}$ then its of integers $R$ is a Dedekind domain, and has unique factorization of ideals into powers of prime ideals. For each prime number $\ell$, you can then look at how $\ell R$ decomposes and the prime factors of this ideal are exactly the primes of $\mathfrak{q}\subset R$ such that $\mathfrak{q}\cap \mathbb{Z} = (\ell)$. If $F$ is Galois over $\mathbb{Q}$, then $\mathrm{Gal}(F/\mathbb{Q})$ acts transitively on the set of prime ideals lying over $\ell$. 
What's the situation if we look at the ring $R'$ of all algebraic integers in $\overline{\mathbb{Q}}$? I read somewhere else that $R'$ is a non-noetherian domain of Krull dimension $1$ and it is integrally closed.
I mean to ask, 


*

*If $\ell$ is a rational prime, are there only finitely many prime ideals of $R'$ lying over $\ell$?

*Does $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ act transitively on the set of such primes?


Better yet, it would be nice to know about a reference for these things.
 A: It's easier to understand prime ideals in the algebraic integers by looking at their intersection with normal number fields :
Let $O$ be the ring of algebraic integers.
If $\mathfrak p$ is a prime ideal in $R$ and $\Bbb Q \subset K \subset \overline{\Bbb Q}$ is a Galois finite extension, then $\mathfrak p_K = \mathfrak p \cap K$ is a prime ideal of $O_K = O \cap K$. Since $\overline{\Bbb Q} = \bigcup K$, we have $\mathfrak p = \bigcup \mathfrak p_K$, and so the data of the $\mathfrak p_K$ is enough to recover $\mathfrak p$.
Conversely, if we are given a family $(\mathfrak p_K)$ for every $K$, such that whenever $K \subset L$, $\mathfrak p_K = \mathfrak p_L \cap K$, then $\mathfrak p = \bigcup \mathfrak p_K$ is a prime ideal of $O$.
The action of the Galois group is also completely determined by the actions of $\mathrm{Gal}(K/\Bbb Q)$ on the $\mathfrak p_K$ : if we have a family $\sigma_K$ such that $\sigma_L|_K = \sigma_K$ when $K \subset L$, then the glueing of the $\sigma_K$ give an automorphism $\sigma$ of $\overline{\Bbb Q}$. Then $(\sigma(\mathfrak p))_K = \sigma_K(\mathfrak p_K)$. 
If it's easier, you can pick a sequence of finite Galois extensions $K \subset K_1 \subset K_2 \subset \cdots$ such that $\bigcup K_n = \overline{\Bbb Q}$, and only use those fields.
Then, to construct an ideal of $O$, you can start from an ideal of $\Bbb Z$ and find a sequence of primes of $O_{K_i}$ lying above one another. Similarly, to construct an automorphism, start from $\mathrm{id}_\Bbb Q$ and find a sequence of automorphisms of $K_i$ extending one another.
From there it's easy to see that if $\mathfrak p_K$ is a prime of $O_K$then there are (uncountably many) primes $\mathfrak p$ above $\mathfrak p_K$, and that the action of $\mathrm{Gal}(\overline{\Bbb Q}/K)$ on them is transitive.
As an explicit example, pick a prime integer $p$ and look at the extensions $ \Bbb Q(\sqrt q)$ where $q$ is prime and $q \equiv 1 \pmod p$. For every $q$, $q$ is a square mod $p$, so $(p)$ splits in two ideals in $\Bbb Q(\sqrt q)$. And for any finite family $(q_1,\ldots,q_n)$, there are $2^n$ prime ideals above $p$ in the composition $\Bbb Q(\sqrt {q_1}, \ldots, \sqrt{q_n})$. Looking at the infinite extension $\Bbb Q(\sqrt{q_i})$, we get uncountably many ideals above $(p)$.
