I am trying to prove 2 statements:
- $\bar{\mathbb{Q}} _\mathbb{R}\subset \mathbb{R}$
- $\bar{\mathbb{R}} _\mathbb{C}=\mathbb{C}$
For (1), I have come up with a few examples such as $\sqrt{2},\sqrt{3}\in\mathbb{R}$ which are algebraic in $\mathbb{Q}$, since we have $f=X^2-2,g=X^2-3\in\mathbb{Q}[X]$ and $f(\sqrt{2})=0$ and $g(\sqrt{3})=0$.
Then, we see that $a+b,a-b,ab\in{\bar{\mathbb{Q}}}_\mathbb{R}$ and $\frac{a}{b}\in{\bar{\mathbb{Q}}}_\mathbb{R}$ for $b\neq0$.
Is this enough to claim that $\bar{\mathbb{Q}} _\mathbb{R}$ is a subfield of $\mathbb{R}$? I feel like this is not enough since I have not explored every element of $\mathbb{R}$ which are algebraic in $\mathbb{Q}$.
For (2), how do I show that the inclusion works both ways, ${\bar{\mathbb{R}}}_\mathbb{C}\subset\mathbb{C}$ and ${\bar{\mathbb{R}}}_\mathbb{C}\supset\mathbb{C}$ such that LHS and RHS are equal, i.e. ${\bar{\mathbb{R}}}_\mathbb{C}=\mathbb{C}$?
NB: By $\bar{K}_F$, I mean algebraic closure of $K$ in $F$
Since $\overline{K}_F$ is the algebraic closure of $K$ in $F$; so it's obvious that $K \subset F$.
So for (2), to prove $\overline{\mathbb{R}}_\mathbb{C} = \mathbb{C}$; all we need to prove is that $\mathbb{C} \subset \overline{\mathbb{R}}_\mathbb{C}$.
Take $a + bi \in \mathbb{C}$; and it's easy to see that this is a solution to the equation $0 = x^2 - 2ax + a^2 + b^2 \in \mathbb{R}[x]$
For (1), https://en.wikipedia.org/wiki/Transcendental_number
It's a well known fact that $\pi \in \mathbb{R}$ is not a solution to any equations with rational coefficient; hence $\pi \notin \overline{\mathbb{Q}}_\mathbb{R}$.
So $\overline{\mathbb{Q}}_\mathbb{R} \subsetneq \mathbb{R}$.