Does $\mathbb{Q}[x_1,x_2\ldots]$ embed inside of $\mathbb{R}$?

Question

It's known that $\pi$ is transcendental over $\mathbb{Q}$, so $\mathbb{Q}[\pi]\cong\mathbb{Q}[x]$.

Is there a way to explicitly construct $a_1,a_2,\ldots\in\mathbb{R}$ such that $\mathbb{Q}[a_1,a_2,\ldots]\cong\mathbb{Q}[x_1,x_2,\ldots]$?

("construct" either from scratch or from knowing something like $\pi$ is transcendental)

I was trying to use the fact that any algebraic function involving just rationals and $\pi$ is still transcendental over $\mathbb{Q}$. But if you pick the $a_i$ like this, then you get an algebraic relation! E.g., if $a_i=\sqrt[i+1]{\pi}$, then $a_i^{i+1}-a_j^{j+1}=0$.

Any other ideas?

Answer 1

The Lindemann-Weierstrass Theorem states

if $\alpha_1, \dots, \alpha_n$ are lineararly independent algebraic numbers, then $e^{\alpha_1},\dots, e^{\alpha_n}$ are algebraically independent transcendental numbers.

For instance, $e^{\sqrt 2}, e^{\sqrt 3}, e^{\sqrt 5}, \dots$ and so on for all prime numbers is a countable algebraically independent set.

Answer 2

Trevor Gunn gave an explicit construction using the Lindermann-Weierstrass Theorem. Inductively constructing $a_1, a_2, \dots$ using a counting argument works as well, although you may not consider that an explicit construction.

Because ${\mathbb Q}[a_1, \dots, a_n] \subseteq {\mathbb R}$ is countable, its algebraic closure (inside ${\mathbb C}$) is countable as well and hence $\overline{{\mathbb Q}[a_1, \dots, a_n]} \cap {\mathbb R}$ is countable as well. Now pick a new element $a_{n+1} \in {\mathbb R}$ that is not yet in $\overline{{\mathbb Q}[a_1, \dots, a_n]} \cap {\mathbb R}$.