$\mathbb{Q}(\zeta_m)=\mathbb{Q}(\zeta_{2m})$

Question

I want to show that if $m$ is odd, $\mathbb{Q}(\zeta_m)=\mathbb{Q}(\zeta_{2m})$, where $\zeta_m$ is the $m$-th root of unity, $\zeta_m=e^{2\pi i/k}$.

This is stated without proof in all the textbooks I've looked, so I guess it can't be very hard.

I'm able to prove it using the fact that $\Phi_{2n}(x)=\Phi_{n}(-x)$, but I'm looking for a more direct proof.

Thanks in advance for any help.

Answer 1

If $m$ is odd, let $\zeta_m$ be a primitive $m^{\mathrm{th}}$ root of unity. Then $(-\zeta_m)^m = (-1)^m = -1$, so that $|-\zeta_m| = 2m$ and thus $-\zeta_m$ is a primitive $2m^{\mathrm{th}}$ root of unity.