why $\zeta_n = e^{2\pi i / n}$, why for p prime divides n, $\mathbb{Q}(\zeta_n) \supset \mathbb{Q}(\zeta_p)$?

Question

When I follow the argument,

($\zeta_n = e^{2\pi i / n}$,) if p prime divides n, $\zeta_p^n = 1$, hence $\mathbb{Q}(\zeta_n) \supset \mathbb{Q}(\zeta_p)$

I did not get the point why we can conclude the containing relationship. Can someone help and explain it more in details, thanks.

Answer 1

I suppose if you know that $x^n-1$ splits in $\mathbb Q(\zeta_n)$ (which requires proof,) the we could say:

Since $\zeta_p^n=1,$ you have $\zeta_p\in \mathbb Q(\zeta_n).$

More generally, $k(\alpha)\subseteq k(\beta)$ if and only if $\alpha\in k(\beta).$

It is much easier to show that $\zeta_p\in\mathbb Q(\zeta_n)$ if $p\mid n.$ Namely: ' $$\zeta_p = \zeta_n^{n/p}.$$