Let $d\in \mathbb Q$ , to prove $\mathbb Q(\sqrt d) \subseteq \mathbb Q(e^{2i\pi/n})$ for some positive integer $n$ (without Kronecker-Weber) Let $d\in \mathbb Q$ , how to show that there exist $n \in \mathbb N$ such that $\mathbb Q(\sqrt d) \subseteq \mathbb Q(e^{2i\pi/n})$ ?
NOTE : I want to do this using basic field extension and Galois theory , without invoking Kronecker-Weber theorem . If possible I would also want to avoid Gauss sum or Legendre symbol
Please help . Thanks in advance 
 A: I will assume that $d \in \Bbb N_{>0}$ (this will cover all the quadratic extensions of $\Bbb Q$). Looking at the decomposition of $d$ into prime factors, we know that $\Bbb Q(\sqrt d)$ is the compositum of some fields $\Bbb Q(\sqrt p)$, where $p$ is either prime or $-1$.
It is therefore sufficient to prove the claim for the fields $\Bbb Q(\sqrt p)$ (where $p$ is either prime or $-1$), since a compositum of cyclotomic fields is again a cyclotomic field.

First of all, the case $p=-1, K=\mathbb{Q}(i) = \mathbb{Q}(\zeta_4)$ is trivial.
Secondly, consider $K=\mathbb{Q}(\sqrt{2})$. We have
$\zeta_8 = \dfrac{\sqrt{2}}{2}(1+i),$ so that
$$\sqrt{2} = \dfrac{2\zeta_8}{1+\zeta_8^2}$$
and $K \subset \mathbb{Q}(\zeta_8)$.

Finally, let $p>2$ be an odd prime and consider $K = \mathbb{Q}(\sqrt{p})$. The discriminant of the $p$-th cyclotomic polynomial is
$$\Delta := \mathrm{disc}(\Phi_p) = \prod_{1 \leq i < j \leq p-1} (\zeta_p^i - \zeta_p^j)^2 = (-1)^{\frac{p-1}{2}} p^{p-2} \in \mathbb{Q}(\zeta_p).$$
This yields
$$\sqrt{p} = \dfrac{\sqrt{(-1)^{\frac{p-1}{2}} \Delta}}{ p^{\frac{p-3}{2}} } \in \mathbb{Q}(i,\zeta_p) = \mathbb{Q}(\zeta_{4p}).$$
Hence, we get $K \subset \mathbb{Q}(\zeta_{4p})$.
