Let $\mathbb{Q}$ be the field of rational numbers, and let $\overline{\mathbb{Q}}$ be its algebraic closure. Assume $\overline{\mathbb{Z}}$ is the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$, my question is that, what can we say about the integral domain $\overline{\mathbb{Z}}$?


Explicitly:


(1) Is it noetherian? (Hardly I think)


(2) How do we describe the nonzero prime ideals of $\overline{\mathbb{Z}}$ ? Are they maximal ? What are the residue fields? (I guess the residues are algebraic closure of finite field)


(3) Is it factorial?


(4) What is $\overline{\mathbb{Z}}^{\times}$ as an abstract group?


For (4), Obviously it contains all roots of unity, and as an abstract group it is isomorphic to $\mathbb{Q}/\mathbb{Z}$. Quotient it out, it becomes uniquely divisible, and hence an $\mathbb{Q}$-vector space. Can we find an explicit basis of it?


All these questions are clear for the integer ring $\mathcal{O}\_K$ of a number field $K$, from algebraic number theory. And it is obvious that $\overline{\mathbb{Z}}$ is the direct limit of the integer ring of number fields, i.e $\overline{\mathbb{Z}}=\varinjlim\_{K}\mathcal{O}\_K$. But I am not quite familiar with the property of direct limit of rings. Can anyone give me some detail?


Additionally, one can ask the following question:


Consider $\overline{\mathbb{Q}}^{\times}$. It can be split into (not canonically) a product of a torsion part $\overline{\mathbb{Q}}^{\times}\_{tor}$ and a free part $\overline{\mathbb{Q}}^{\times}/\overline{\mathbb{Q}}^{\times}\_{tor}$ . The torsion part are exactly roots of unity, and is $\mathbb{Q}/\mathbb{Z}$. The quotient part becomes a $\mathbb{Q}$ vector space, too. What can we say about it?



The integral closure $\overline{\mathbb{Z}}$ and the group $\overline{\mathbb{Z}}^{\times}$
For (3), the answer is no, because there is no irreducible element in $\overline{\mathbb{Z}}$: for example, every element is a square.


Similarly for (1), the answer is no because there are infinite ascending chains of ideals: for example, $(x) \subset (x^{1/2}) \subset (x^{1/4}) \subset \cdots$, where $x$ is not a unit (e.g. $x=2$) and $\{x^{1/2^n}\}$ is a compatible system of $2^n$-th roots of $x$ in $\overline{\mathbb{Z}}$.


For (4), the group $\overline{\mathbb{Z}}^\times$ is divisible, and when we quotient by the torsion, we get a uniquely divisible group $P$, hence a $\mathbb{Q}$-vector space. The dimension must be countable.


**EDIT.** As Laurent Moret-Bailly points out, the exact sequence
\begin{equation\*}
0 \to \overline{\mathbb{Z}}^\times\_{\mathrm{tors}} \to \overline{\mathbb{Z}}^\times \to P \to 0
\end{equation\*}
splits because $\overline{\mathbb{Z}}^\times\_{\mathrm{tors}}$ (the group of all roots of unity) is isomorphic to $\mathbb{Q}/\mathbb{Z}$, which is a divisible abelian group, hence is injective.


Similarly, consider the exact sequence
\begin{equation\*}
(\*) \qquad 0 \to \overline{\mathbb{Z}}^\times \to \overline{\mathbb{Q}}^\times \to P' \to 0.
\end{equation\*}
Then $P'$ is a $\mathbb{Q}$-vector space of countable dimension. Moreover $\overline{\mathbb{Z}}^\times \cong (\mathbb{Q}/\mathbb{Z}) \oplus P$ is injective (this can be proved using Baer's criterion). Hence the sequence $(\*)$ splits, and we get isomorphisms
\begin{equation\*}
\overline{\mathbb{Z}}^\times \cong \frac{\mathbb{Q}}{\mathbb{Z}} \oplus \mathbb{Q}^{(\mathbb{N})} \qquad \textrm{and} \qquad \overline{\mathbb{Q}}^\times \cong \overline{\mathbb{Z}}^\times \oplus \mathbb{Q}^{(\mathbb{N})}
\end{equation\*}
A priori these isomorphisms are totally non-canonical.


**EDIT 2.** As pointed out in the comments, I did not prove that the dimension of $P$ is infinite. This is a consequence of Dirichlet's unit theorem (see the comments), but we can give an elementary proof. Namely, consider the algebraic units $u=t+ \sqrt{t^2+1}$ where $t \geq 1$ is an integer. They have infinite order since the only real roots of unity are $\pm 1$, and obviously $u \geq 1+\sqrt{2}$. Now suppose that $t\_1, \ldots,t\_n$ are integers such that the integers $d\_i = t\_i^2+1$ are pairwise coprime. Then the fields $\mathbb{Q}(\sqrt{d\_i}) = \mathbb{Q}(u\_i)$ are linearly disjoint (it is not too difficult to prove this by induction on $n$, see [here](https://math.stackexchange.com/questions/30687/the-square-roots-of-different-primes-are-linearly-independent-over-the-field-of)). In particular $u\_n \notin \mathbb{Q}(u\_1,\ldots,u\_{n-1})$, and in fact $u\_n^a$ cannot belong to $\mathbb{Q}(u\_1,\ldots,u\_{n-1})$ for any $a$, since otherwise $u\_n^a$ would be an algebraic unit in $\mathbb{Q}$, hence equal to $\pm 1$. By induction, this shows that the units $u\_1,\ldots,u\_n$ are independent. Finally, to find the integers $t\_n$, we can define them inductively with $t\_1=1$ and $t\_{n+1} = d\_1 \cdots d\_n$ for any $n \geq 1$.


