13

$\overline{\Bbb{Q}}$ is the field of algebraic numbers, let $E$ be the set of embeddings $\overline{\Bbb{Q}}\to \Bbb{C}$ and consider the following norm on $\overline{\Bbb{Q}}$ $$\|\alpha\|=\sup_{\sigma\in E} |\sigma(\alpha)|$$ making it a topological field. The completion for that norm is a ring containing $\overline{\Bbb{Q}}\otimes_\Bbb{Q}\Bbb{R}$ (including some zero divisors such as $(1\otimes \sqrt2)^2=(\sqrt2\otimes 1)^2$) as well as things such as $$\sum_{n\ge 1} 2^{-n} \sqrt{n}$$

Is there a simple description of this completion?

reuns
  • 79,880
  • How can I see that $E$ contains $\overline{\mathbb{Q}}\otimes_\mathbb{Q} \mathbb{R}$? I think $E$ should be a subring of $\overline{\mathbb{Q}}\otimes_\mathbb{Q} \mathbb{C}$. – Ben Nov 19 '22 at 23:05
  • 1
    The sequence $a_N= \sum_{n=1}^N 2^{-n} \sqrt{n}\in \overline{\Bbb{Q}}$ converges for the norm and the limit $\lim_{N\to \infty} a_N$ is not in $\overline{\Bbb{Q}}\otimes \Bbb{C}$. The completion of a number field $K\cong \Bbb{Q}[x]/(f)$ for the norm is $K \otimes \Bbb{R}\cong \Bbb{R}[x]/(f)$, the $K \otimes \Bbb{R}$ for various $K$ give that $\overline{\mathbb{Q}}\otimes_\mathbb{Q} \mathbb{R}\subset E$. @BenS. – reuns Nov 19 '22 at 23:22
  • Thank you @reuns , I think I get where I made a mistake, although I'll have to think about why the completion of $K$ is $K\otimes \mathbb{R}$. However couldn't you make a colimit argument to argue that $E=\overline{\mathbb{Q}}\otimes \mathbb{R}$? Completion should commute with colimits since its left-adjoint and $\overline{\mathbb{Q}}$ is a colimit of number fields. I guess there a some details that need to be checked... – Ben Nov 19 '22 at 23:48
  • $\overline{\mathbb{Q}}\otimes \mathbb{R}$ is an infinite dimensional $\Bbb{R}$-algebra but every element lives in a finite dimensional $\Bbb{R}$-algebra, whereas $\lim_{N\to \infty} a_N$ doesn't @BenS. – reuns Nov 20 '22 at 00:45
  • There’s a natural isometric $\mathbb{Q}$-embedding $\overline{\mathbb{Q}} \rightarrow {f \in C^0(E,\mathbb{C}),, f(\overline{s})=\overline{f(s)}}$ of algebras… To show that it’s a completion (which solves your problem – but I’m not quite sure it’s correct), all we need to do is show that it has dense image. I wanted to use Stone-Weierstrass – point separation is easy, but the image doesn’t seem stable under complex conjugation. Maybe we can still prove dense image using Riesz-Markov – ie if $\mu$ is a signed measure on $E$ annihilating every $e_z$, then it anti-commutes with conjugation. – Aphelli Nov 21 '22 at 23:17
  • @Aphelli Looks great – reuns Nov 22 '22 at 08:29
  • @Aphelli The main interest in this ring is to apply automorphisms of $\in G_\Bbb{Q}$ to things like $\exp(\sqrt2)$, and if $\exp(\sqrt2)$ is in fact meant to be a continuous function from $G_{\Bbb{Q}(i)}$ to $\Bbb{C}$ then it becomes much less mysterious why we can apply automorphisms $\in G_{\Bbb{Q}(i)}$ to it. – reuns Nov 22 '22 at 15:58

1 Answers1

4

$E$ is endowed with a natural profinite topology (homeomorphic to $G_{\mathbb{Q}}$). Fix a primitive $4$-th root of unity $\zeta \in \overline{\mathbb{Q}}$ and let $E’=\{s \in E,\, s(\zeta)=i\}$, it’s profinite as well (homeomorphic to $G_{\mathbb{Q}(i)}$).

I claim that your space is exactly the algebra $V=C^0(E’,\mathbb{C})$ for the $\|\cdot\|_{\infty}$ norm.

Indeed, let $z \in \overline{\mathbb{Q}}$. Define as $e_z(\sigma)=\sigma(z)$ for $\sigma \in E’$. This defines a map $F: z \in \overline{\mathbb{Q}} \longmapsto e_z \in V$.

Indeed, if $z \in K$, then $e_z(\sigma)$ depends only on $\sigma_{|K}$, which is exactly what it means for $e_z$ to be continuous.

By definition, $\|F(z)\|_{\infty}=\|z\|$, so that $F$ is an isometric embedding of algebras. Thus it extends to an isometric embedding of algebras $\overline{F}: C \rightarrow V$, where $C$ is the completion that we’re interested in. The goal is to show that $\overline{F}$ is an isomorphism.

Write $C_1=\overline{F(C)} \subset V$. Then $\mathbb{R} \subset C_1$, and $e_{\zeta}$ is the constant function $i$, so $C_1$ is a $\mathbb{C}$-subalgebra of $V$.

The problem is solved if we show that $C_1$ contains all the locally constant functions of $V$.

So let $K \supset \mathbb{Q}(\zeta)$ be a number field, and consider the $\psi: E’ \rightarrow \mathbb{C}$ such that $\psi(s)=\psi(s’)$ if $s_{|K}=s’_{|K}$. The $\mathbb{C}$-vector space $\Psi_K$ of such $\psi$ has dimension exactly $[K:\mathbb{Q}(\zeta)]$.

Now, it contains the $\mathbb{R}$-vector space $E_K$ generated by the $e_z$ with $z \in K$. But we know from classical “geometry of numbers” that this space has dimension $[K:\mathbb{Q}]$, ie $\dim_{\mathbb{R}}{\Psi_K}=\dim_{\mathbb{R}}{E_K} <\infty$, and $E_K \subset \Psi_K$, so $\Psi_K = E_K \subset C_1$. QED.

Aphelli
  • 37,929