So after the discussion with Keenan above here is how we settle the case when $m = p^r$. Now by definition the $m$ - th cyclotomic polynomial
$\Phi_m(x) = \prod_{1\leq k \leq m-1 : (m,k) = 1}(x- \zeta_m^k) $
and since $\zeta_m$ is a root of this polynomial this means that $[\Bbb{Q}(\zeta_m):\Bbb{Q}] \leq \varphi(m)$, the degree of the $m$- th cyclotomic polynomial. Now because $\Bbb{Q}(\zeta_m)$ is a Galois extension we have that the ramification indices of all primes that lie above $p$ in $\mathcal{O}_{\Bbb{Q}(\zeta_m)}$ are all equal, similarly for the inertia degrees. So we can say that
$[\Bbb{Q}(\zeta_m):\Bbb{Q}] = efg$
where $e$ is the ramification index of $p$, $f$ the inertia degree and $g$ the number of distinct prime ideals in the prime factorisation of $(p)$ in $\mathcal{O}_{\Bbb{Q}(\zeta_m)}$. Now it is clear that we have inclusions $\Bbb{Z} \subseteq \Bbb{Z}[\zeta_m] \subseteq \mathcal{O}_{\Bbb{Q}(\zeta_m)}.$
Since the ramification index is multiplicative in towers, we have
$e = e(Q|(1-w))e((1-w)|p) = e(Q|(1-w))\varphi(m)$
where $Q$ is any prime ideal of $\mathcal{O}_{\Bbb{Q}(\zeta_m)}$ lying over $(1-w)$ and hence over $p$. This means that the number $e$ and hence the number $[\Bbb{Q}(\zeta_m) : \Bbb{Q}] \leq \varphi(m)$ and so we conclude that
$[\Bbb{Q}(\zeta_m) : \Bbb{Q}] = \varphi(m).$
Now for the general case. By the counting formula that I invoked in one of my comments to Keenan's answer above, it is enough to prove that
- $\Bbb{Q}(\zeta_{p^r}) \cap \Bbb{Q}(\zeta_{l^s}) = \Bbb{Q}$.
- $\Bbb{Q}(\zeta_{p^r})\Bbb{Q}(\zeta_{l^s}) = \Bbb{Q}(\zeta_{p^rl^s}).$
I will only prove the first of these for the second is a standard fact concerning cyclotomic extensions. By (b) to prove the first of these we need to show that if $p$ is a prime not equal to $l$, then $(p)$ is unramified in $\Bbb{Q}(\zeta_{l^s})$ for any $s \geq 1$. To do this we recall the following fact from algebraic number theory:
Let $p$ be a prime in $\Bbb{Z}$ and suppose $p$ is ramified in a number ring $R$. Then $p |\textrm{disc}(R)$.
We already know that $\textrm{disc}(\zeta_{l^s})$ divides $l^{s\varphi(l^s)}$ and so $\textrm{disc}(\zeta_{l^s})$ is a power of $l$. Since $p$ does not divide any power of $l$ by the theorem above $p$ is unramified in $\Bbb{Q}(\zeta_{l^s})$.