Cyclotomic Fields are equal? We know that the $m$th cyclotomic number field is given by $\mathbb{Q}(\xi_m)$, where $\xi_m$ is an $m$th primitive root of unity. We know also that if $\omega_m$ is another primitive root of unity, then $\mathbb{Q}(\xi_m)$ and $\mathbb{Q}(\omega_m)$ are isomorphic, since they have the same minimal polynomial $\Phi_m(X)$. My question is:
Can we say that $\mathbb{Q}(\xi_m)=\mathbb{Q}(\omega_m)$ ? (so that the definition of $m$'th cyclotomic number field is well defined)?
 A: There is a subtlety here: are you trying to compare two subfields of a common field or two fields that are built separately? 
If you have two primitive $m$th roots of unity in the same field of characteristic $0$ then they generate the same cyclotomic fields over $\mathbf Q$. But if your $\xi_m$ and $\omega_m$ are not in the same field then the fields they generate over $\mathbf Q$ are isomorphic but can't be equal.
For example, $\mathbf Q(i)$ inside $\mathbf C$ and $\mathbf Q[x]/(x^2+1)$ are isomorphic but definitely are not equal.
A: Yes, you can. Fix $n$ and positive integers $a,b$ relatively prime to $n$. Then if $\zeta_n = e^{2\pi i \frac{1}{n}}$, $\zeta_n^a$ and $\zeta_n^b$ are primitive $n^{th}$ roots of unity. We can solve the congruence $ax \equiv b \pmod{n}$. Then $(\zeta_n^a)^x = \zeta_n^{ax} = \zeta_n^b$, which means that $\zeta_n^b \in \mathbb{Q}(\zeta_n^a)$ and thus $\mathbb{Q}(\zeta_n^b) \subseteq \mathbb{Q}(\zeta_n^a)$. Of course, symmetry gives $\mathbb{Q}(\zeta_n^a) \subseteq \mathbb{Q}(\zeta_n^b)$. 
