Is $\Bbb{R}$ finitely generated over the algebraics?

Question

Let $K$ be the field of all algebraic numbers. Then does there exist an algebraically independent set of numbers $\pi_1, \dots, \pi_n$ such that $\Bbb{R} = K(\pi_1, \dots, \pi_n)$?

I don't know where to begin on this one or if it's even possible with my limited tools to solve it. Any ideas?

One thing I know is that if $r \notin L = K(\pi_1, \dots, \pi_n)$, then the minimal polynomial of $r$ over $K$ must have at least one transcendental coefficient.

Answer 1

Show that if $K$ is a countable field, then any finitely generated extension $K(x_1,\dots,x_n)$ is also countable.

One way to do this is to first show that the ring $K[x_1,\dots,x_n]$ is countable, and then use the fact that $K(x_1,\dots,x_n)$ is the fraction field of $K[x_1,\dots,x_n]$.

Answer 2

Hint: $\mathbf R$ is uncountable.