Criterion for subfield of $ \mathbb{C} $ to be dense

Question

Question: Is it true that a subfield $ K $ of $ \mathbb{C} $ is dense if and only if the roots of unity in $ K $ are dense in the unit circle?

Context: I was thinking about the infinite degree algebraic extension $ K:=\mathbb{Q}(\zeta_p, \zeta_{p^2},\zeta_{p^3},\dots ) $ adjoin all the $ p^n $ roots of unity for a fixed $ p $. It seems like $ K $ should be dense in $ \mathbb{C} $.

Answer 1

Being a dense subfield of $\mathbb{C}$ is not very difficult. Every complex embedding of a number field $K \to \mathbb{C}$ (i.e. with image not contained in the real line) is dense in the Euclidean topology of $\mathbb{C}$, but $K$ always has finitely many roots of unity. For example, $\mathbb{Q}(i)$ is dense in $\mathbb{C}$, but the only roots of unity it contains are $\{\pm 1, \pm i\}$, which is not dense in the unit circle.

Conversely, any subfield of $\mathbb{C}$ containing any root of unity besides $\pm 1$ is dense in $\mathbb{C}$, since such a field necessarily contains a cyclotomic number field $\mathbb{Q}(\zeta_m)$, and this is dense in $\mathbb{C}$ already.

Answer 2

$\mathbb Q(i)\subset\mathbb C$ is dense. This follows from observing $\mathbb Q\subset\mathbb R$ is dense (essentially by definition).

Answer 3

Dense subfields of $\Bbb{C}$ need not contain any roots of unity besides $z = \pm 1$.
Every subfield $F \subset \Bbb{C}$ contains $\Bbb{Q}$, so a necessary and sufficient condition for a subfield $F$ to be dense in $\Bbb{C}$ is that $F$ contain at least one complex number $\alpha$ with nonzero imaginary part.
An explicit example would be $F := \Bbb{Q}( \pi i)$, which is dense in $\Bbb{C}$ but cannot contain any complex roots of unity, since $\pi$ is transcendental.