In an answer to another question I used the fact that $\mathbb{Q}(\zeta_m)\subseteq \mathbb{Q}(\zeta_n)$ if and only if $m$ divides $n$ (here $\zeta_n$ stands for a primitive $n$th root of unity, Edit: and neither $m$ nor $n$ is twice an odd number; see KCd comments below).
More generally, one can show that $\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}$ when $\gcd(n,m)=1$. The only proof of this fact that comes to mind uses facts about discriminants of cyclotomic extensions, and the fact that every non-trivial number field extension over $\mathbb{Q}$ ramifies at least at one prime (see, for instance, Washington, "Introduction to Cyclotomic Fields", Chapter 2, Proposition 2.4).
Since the original question that I was trying to answer was somewhat elementary, I was left wondering if there are more elementary proofs of the fact
$$\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}, \text{ when } \gcd(n,m)=1. $$
By "elementary proof" here I mean some proof that does not involve algebraic number theory results about discriminants, or ramification of primes in rings of integers of number fields.
Can anyone think of an elementary proof?
Thanks!
Best Answer
$\newcommand{\Q}{\mathbf Q}$This answer assumes that we're willing to use $[\Q(\zeta_n) : \Q] = \varphi(n)$, which is not obvious. A freely available reference is Milne's notes, Lemma 5.9 and Theorem 5.10.
Since $n$ and $m$ are coprime, the proof you gave shows that $\mathbf Q(\zeta_{nm}) = \mathbf Q(\zeta_n, \zeta_m)$, and the totient function satisfies $\varphi(nm) = \varphi(n)\varphi(m)$. Now \[ [\mathbf Q(\zeta_{nm}) : \mathbf Q] = [\mathbf Q(\zeta_n, \zeta_m) : \mathbf Q(\zeta_n)][\mathbf Q(\zeta_n) : \mathbf Q]. \] If we had $\mathbf Q(\zeta_n) \cap \mathbf Q(\zeta_m) \neq \mathbf Q$ then the degree of $\zeta_m$ over $\mathbf Q(\zeta_n)$ would be less than $\varphi(m)$.