If $n$ divides $m$, prove that $\mathbb{Q}(\zeta_{n}) \subset \mathbb{Q}(\zeta_{m})$

Question

If $n$ divides $m$, prove that $\mathbb{Q}(\zeta_{n}) \subset \mathbb{Q}(\zeta_{m})$.

If $n$ divides $m$, so $m = nk$ and $\varphi(m) = \varphi(n)\varphi(k)(k,n)/\varphi((k,n))$, then $\varphi(n)$ divide $\varphi(m)$ and thus, $[\mathbb{Q}(\zeta_{n}):\mathbb{Q}]$ divides $[\mathbb{Q}(\zeta_{m}):\mathbb{Q}]$. But, this doesn't ensure that $\mathbb{Q}(\zeta_{n}) \subset \mathbb{Q}(\zeta_{m})$. How do I do this?

Answer 1

I don't think you even need to talk about field extensions answer this. If $\omega$ is a primitive $n$^th root of unity and $n$ divides $m$, then it'll be an $m$^th root of unity too since $$\omega^m = \omega^{nk} =(\omega^n)^k = 1^k = 1$$ so $\omega$ must live in the $m$th cyclotomic field, which contains all the $m$^th roots of unity.