Clarification on proof that $\mathbb Q (\zeta _ m) \cap \mathbb Q(\zeta_n) = \mathbb Q$ when $\gcd(m,n) = 1$ In this proof of the result in the title, Dylan says that because of the Tower Law relation:
$$\left[\mathbb Q(\zeta_{nm}) : \mathbb Q\right] = \left[\mathbb Q(\zeta_{m},\zeta_n) : \mathbb Q(\zeta_n)\right]\left[\mathbb Q(\zeta_n):\mathbb Q\right]$$
Then if $\mathbb Q(\zeta_n) \cap \mathbb Q(\zeta_m) \neq \mathbb Q$, we must have that $\left[\mathbb Q(\zeta_n, \zeta_m):\mathbb Q(\zeta_n)\right] < φ(m)$.
However, I really don't see how he made this deduction, or even how this relation is relevant. As far as I'm aware, because we don't know what $\left[\mathbb Q(\zeta_n):\mathbb Q(\zeta_n)\cap \mathbb Q(\zeta_m)\right]$ is, we can't conclude what Dylan did.
If anyone could help explain this to me that would be great thank you.
 A: If $L/k$ is a field extension and $x \in L$ is algebraic over $k$, then I denote by $\pi_{x, k}$ its minimal polynomial.  
Let's get back to our problem:  


*

*First, notice that since $\mathbb{Q}(\zeta_{n}) \cap \mathbb{Q}(\zeta_{m}) \subset \mathbb{Q}(\zeta_{n})$, $\pi_{\zeta_{m}, \mathbb{Q}(\zeta_{n})}$ divides $\pi_{\zeta_{m}, \mathbb{Q}(\zeta_{n}) \cap \mathbb{Q}(\zeta_{m})}$ in $\mathbb{Q}(\zeta_{n})[X]$ and hence, $$[\mathbb{Q}(\zeta_{n}, \zeta_{m}) : \mathbb{Q}(\zeta_{n})] \leq [\mathbb{Q}(\zeta_{m}) : \mathbb{Q}(\zeta_{n}) \cap \mathbb{Q}(\zeta_{m})]$$  

*On the other hand, $$[\mathbb{Q}(\zeta_{m}) : \mathbb{Q}] = [\mathbb{Q}(\zeta_{m}) : \mathbb{Q}(\zeta_{n}) \cap \mathbb{Q}(\zeta_{m})] .[\mathbb{Q}(\zeta_{n}) \cap \mathbb{Q}(\zeta_{m}) : \mathbb{Q}]$$  


Therefore, if $\mathbb{Q}(\zeta_{n}) \cap \mathbb{Q}(\zeta_{m}) \ne \mathbb{Q}$, then (by the second point) $[\mathbb{Q}(\zeta_{m}) : \mathbb{Q}(\zeta_{n}) \cap \mathbb{Q}(\zeta_{m})] < \varphi(m)$ and the desired result follows directly from the first point.
A: Let $L_1, L_2$ be any two Galois extensions of a base field $K$, and let $M = L_1 L_2$. Then there is a canonical identification $\mathrm{Gal}(M/L_1) \cong \mathrm{Gal}(L_2/L_1 \cap L_2)$. You should try to prove this for yourself if you haven't seen it before, but a proof can be found in Keith Conrad's notes here.
Applying this to $K = \mathbf{Q}, L_1 = \mathbf{Q}(\zeta_n), L_2 = \mathbf{Q}(\zeta_m)$, we get $$\mathrm{Gal}(\mathbf{Q}(\zeta_n, \zeta_m)/ \mathbf{Q}(\zeta_n)) \cong \mathrm{Gal}(\mathbf{Q}(\zeta_m) / \mathbf{Q}(\zeta_n) \cap \mathbf{Q}(\zeta_m)).$$
In particular, we have an equality when comparing the degrees of the relative extensions. But $[\mathbf{Q}(\zeta_m): \mathbf{Q}] = \varphi(m)$, and so if $\mathbf{Q}(\zeta_m) \cap \mathbf{Q}(\zeta_n) \neq \mathbf{Q}$ we must have the degree of $\mathbf{Q}(\zeta_m)$ over the intersection (and hence $[\mathbf{Q}(\zeta_m, \zeta_n) : \mathbf{Q}(\zeta_n)]$ as well) strictly smaller than $\varphi(m)$.
