Say that an algebraic number $\alpha\in{\mathbb C}$ is totally real iff ${\mathbb Q}[\alpha]$ is a totally real number field. The totally real numbers obviously form a subfield of $\mathbb R$, which I will denote by $\mathbb T$.
I ask myself several questions on the field ${\mathbb F}={\mathbb T}[i]$ : is it equal to ${\mathbb C}$ ? What is the intersection ${\mathbb F}\cap {\mathbb R}$ ?
The famous theorem of Artin&Schreier is obviously related to this, but I'm not sure how it might help here.
On
You can't hope for it to equal $\mathbb{C}$: the most you could possibly hope for is the algebraic closure of $\mathbb{Q}$.
However, you don't get that: the algebraic closure of a field is a finite-degree extension if and only if the field itself is either real closed or is already algebraically closed.
Every real closed field is ordered and every positive number has a square root. $\mathbb{T}$ cannot be real closed: if $\alpha$ is any nonzero element that is neither totally positive nor totally negative, then neither $\sqrt{\alpha}$ nor $\sqrt{-\alpha}$ are totally real, and thus neither is an element of $\mathbb{T}$.
On
Here is another approach to showing $\mathbb T(i) \cap \mathbb R = \mathbb T$: we have $$ \mathbb T \subset \mathbb T(i) \cap \mathbb R \subset \mathbb T(i) $$ and the second containment is strict since $i \in \mathbb T(i)$ and $i \not\in \mathbb T(i) \cap \mathbb R$. Since $[\mathbb T(i):\mathbb T] = 2$, the middle field must be the bottom field: $\mathbb T(i) \cap \mathbb R = \mathbb T$.
$${\bf F}\cap{\bf R}={\bf T}$$ because $${\bf F}=\{\,a+bi:a,b\in{\bf T}\,\}$$ Since ${\bf T}\subset{\bf R}$, an element $a+bi$ of $\bf F$ is in $\bf R$ if and only if $b=0$, so said element of $\bf F$ is already in $\bf T$.