Infinite field extension but algebraic subset

Question

Is there a subset $ S \subseteq \mathbb{C}$ so that all $s \in S$ are algebraic over $\mathbb{Q}$ and $\left[\mathbb{Q}[S] \colon \mathbb{Q} \right]=\infty$?

$\mathbb{Q}[S]$ is defined as $\mathbb{Q}[S]=\left\lbrace f(x_1,\dotsc,x_n); n \in \mathbb{N}, f \in \mathbb{Q}[X_1,\dotsc,X_n], x_1,\dotsc,x_n \in S \right\rbrace$.

Answer 1

Take, for instance, $S = \{ \sqrt[p]{2} \mid p \in {\mathbb N}, p \geq 2\}$. Each ${\mathbb Q}[\sqrt[p]{2}]$ is a subfield of ${\mathbb Q}[S]$ and $[{\mathbb Q}[\sqrt[p]{2}] : {\mathbb Q}] = p$; therefore $[{\mathbb Q}[S]:{\mathbb Q}] \geq p$ for all $p$.

For finite $S$ this doesn't work; then $[{\mathbb Q}[S] : {\mathbb Q}]$ is always finite.

Answer 2

$S$ could be for instance:

• all algebraic numbers
• all roots of unity $e^{2\pi i\Bbb Q}$
• all $\sin$'s or $\cos$'s of $2\pi \Bbb Q$
• all $\sqrt[n]{a}$ for $a\in\Bbb Z$ and $n$ fixed
• all $\sqrt[n]{a}$ for $n\in\Bbb Z$ and $a$ fixed
• ....