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 $\overline{\mathbb{Z}}$?


(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?

  • $\begingroup$ For (3) the answer is no, because there is no irreducible element in $\overline{\mathbb{Z}}$: for example, every element is a square. $\endgroup$ – François Brunault May 7 at 7:30
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    $\begingroup$ It's neither noetherian nor UFD, but it is a Bézout domain. $\endgroup$ – Emil Jeřábek May 7 at 7:32
  • $\begingroup$ @Emil Jeřábek Thanks but why it isn't noetherian? I cannot give an easy explicit example $\endgroup$ – Yuan Yang May 7 at 7:52
  • $\begingroup$ @François Brunault you are right! $\endgroup$ – Yuan Yang May 7 at 7:52
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    $\begingroup$ For one thing, a noetherian Bézout ring is a PID, and therefore UFD, which $\overline{\mathbb Z}$ is not. Explicitly, use François Brunault’s example: for most elements $a$, the prime ideals generated in turn by $a$, $a^{1/2}$, $a^{1/4}$, ... form a strictly increasing chain. $\endgroup$ – Emil Jeřábek May 7 at 8:44

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). 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 $n \geq 1$.

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    $\begingroup$ Since the torsion of $\overline{\mathbb{Z}}^\times$ is divisible, it is an injective $\mathbb{Z}$-module, so $\overline{\mathbb{Z}}^\times$ is in fact isomorphic to $(\mathbb{Q}/\mathbb{Z})\times P$. $\endgroup$ – Laurent Moret-Bailly May 7 at 8:15
  • $\begingroup$ Since $\mathbb Q^{(\omega)}\oplus\mathbb Q^{(\omega)}\simeq\mathbb Q^{(\omega)}$, this makes $\overline{\mathbb Q}^\times\simeq\overline{\mathbb Z}^\times$. $\endgroup$ – Emil Jeřábek May 7 at 11:43
  • $\begingroup$ Actually, why do $P$ and $P'$ have infinite dimension? I can see this is true for $P'$, as already $\mathbb Q^\times/(\mathbb Q^\times\cap\overline{\mathbb Z}^\times)=\mathbb Q^\times/\{1,-1\}\simeq\mathbb Z^{(\omega)}$ has infinite rank. Is there a simple argument for $P$? $\endgroup$ – Emil Jeřábek May 7 at 11:54
  • $\begingroup$ @EmilJeřábek in a number field $K$ the dimension is $r_1+r_2-1$ , where $ r_1$ and $r_2$ are numbers of real/complex places of K. This can be arbitrary large? $\endgroup$ – Yuan Yang May 7 at 12:00
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    $\begingroup$ @YuanYang We cannot expect a basis of $P'$ to be indexed by the prime ideals of $\overline{\mathbb{Z}}$, because this set is uncountable (if you write $\overline{\mathbb{Q}}$ as a tower of number fields, then each time you go up the tower you must choose a prime ideal, and this needs to be done countably many times). But I really wish there is something similar! $\endgroup$ – François Brunault May 8 at 11:04

I think we can describe $P$ a bit more, using Dirichlet’s unit theorem.

Since $\overline{\mathbf{Z}} = \varinjlim_{[K:\mathbf{Q}] < \infty} \mathcal{O}_K$, the same is true for the units. Now Dirichlet tells us that taking logs of the archimedean valuations $v \mid \infty$ of $K$ gives us an embedding:

$$\mathcal{O}_K^\times/\mu_\infty(K) \hookrightarrow \mathcal{L}_K := (\oplus_{v \mid \infty} \mathbf{R}_v)_0$$

Here, the subscript $(\underline{ })_0$ means the subspace where the coordinates sum to zero. Moreover, this induces an isomorphism after extending scalars to $\mathbf{R}$. The image is a discrete subgroup, so it must be a $\mathbf{Z}$-lattice with rank equal to the real dimension of the space on the right, which is $1$ fewer than the number of archimedean valuations.

If $L/K$ is a finite extension, the inclusion $\mathcal{O}_K^\times \hookrightarrow \mathcal{O}_L^\times$ corresponds to the diagonal embedding of $\mathbf{R}_v$ into $\oplus_{v’\mid v} \mathbf{R}_{v’}$ where the $v’$ are the archimedean valuations of $L$ which extend $v$.

Now, direct limits of abelian groups are exact, so taking limits of both sides gives us an embedding $$P = \overline{\mathbf{Z}}^\times/(\mu_\infty) \hookrightarrow \overline{\mathcal{L}} := \varinjlim_K \mathcal{L}_K \subseteq \prod_{v \mid \infty} \mathbf{R}_v$$ where now the $v$ range through the set of archimedean valuations on $\overline{\mathbf{Q}}$, i.e. the set of conjugate pairs of complex embeddings.

The space $\overline{\mathcal{L}}$ is the set of vectors in the infinite product which are fixed by an open subgroup of the absolute Galois group (which acts transitively on the factors by permutation, with the stabilizer of a given $v$ given by the associated complex conjugation) which moreover have “trace” $0$, in the sense that averaging over the finite Galois orbit gives $0$.

A spanning set is given by the countable set $\{e_{K, v_K} - e_{K, w_K}\}$ where $K$ is a number field, $v_K, w_K$ are archimedean places of $K$, and $e_{K, v_K}$ is the vector with $1$’s in every component $\mathbf{R}_v$ such that $v$ restricts to $v_K$ and $0$’s elsewhere. Certainly this set is not linearly independent, but I can’t think of a natural basis (if anyone else can, please share in the comments!).

Since direct limits of abelian groups moreover commute with tensor products, we know that the induced map $P \otimes_{\mathbf{Z}} \mathbf{R} = P \otimes_{\mathbf{Q}} \mathbf{R} \rightarrow \overline{\mathcal{L}}$ is an isomorphism. Thus, the $\mathbf{Q}$-vector space $P$ is a $\mathbf{Q}$-“lattice” inside $\overline{\mathcal{L}}$.

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    $\begingroup$ I roughly have this picture in my mind but you write it out explicitly. I guess following this strategy one can do a similar thing for any non-archimedean places, and combine them together to get the $Adele$ and $Idele$ of $\overline{\mathbb{Z}}$. That's awesome! $\endgroup$ – Yuan Yang May 7 at 13:23
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    $\begingroup$ If $K'/K$ is a finite extension then the norm map provides a splitting of $\mathcal{O}_K^\times \otimes \mathbb{Q} \hookrightarrow \mathcal{O}_{K'}^\times \otimes \mathbb{Q}$ and similarly for $\mathcal{L}_K \hookrightarrow \mathcal{L}_{K'}$, so there is a natural complement. I don't know if there is a natural basis for this complement. But then we must write $\overline{\mathbb{Q}}$ as a increasing union of number fields and this is quite unnatural... $\endgroup$ – François Brunault May 7 at 15:34

The ring $\overline{\mathbb{Z}}$ is called the ring of algebraic integers. You can find information about prime ideals, e.g., in https://math.stackexchange.com/questions/156231/non-zero-prime-ideals-in-the-ring-of-all-algebraic-integers.

  • $\begingroup$ Thanks for your link! I think I can handle prime ideals. What about the structure of $\overline{\mathbb{Z}}^{\times}$? $\endgroup$ – Yuan Yang May 7 at 7:48

(1) No, as Brunault explaned, we may get an infinite series of ideals. In more general setting, if $A$ is an integral domain then any infinitely ramified (Indeed, an integral extension $B$ of $A$ is "infinitely ramified" if there is a tower $A\subsetneq A_1\subsetneq A_2\subsetneq \cdots$ such that $A_{i+1}/A_i$ are ramified) integral extension of $A$ is "not" noetherian. noetherian condition does not hold for almost all infinite integral extensions! In the setting of $A=\mathbb{Z}_p$, "ramified" extension of $A$ is an extension which "changes" the range of valuation of $A$ and infinitely ramified extension of $A$ has the valuation which has dense range on $\mathbb{R}$! (In contrast, the maximal unramified extension of $A$ has the valuation which has the range $\mathbb{Z}$.) The "denseness" determines noetherian condition.

(2) Any nonzero prime ideal is maximal by the fact that $\overline{\mathbb{Z}}$ is an integral domain and Krull dimension 1 and your guess is right.

(3) As Tom De Medts said, every element is a square and as (1), almost all infinite integral extensions of $\mathbb{Z}$ are not factorial.


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