Another completion of $\overline{\Bbb{Q}}$ $\overline{\Bbb{Q}}$ 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?
 A: $E$ 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.
