Q:

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.

