Classification of cyclotomic extenson PS: Oops! This is somewhat duplicate. Thanks for those who made interest for this post. I will delete this post in a few hours. Please cite the following link
Which cyclotomic fields are different?
PS2: Now this post has a one correct answer. Should I delete this post?
Let $\zeta_n \in \mathbb{C}$ be a primitive $n^{\textrm{th}}$ root of unity and call $\mathbb{Q}[\zeta_n]$ the $n^{\textrm{th}}$ cyclotomic extension of $\mathbb{Q}$.

Let $m$ and $n$ be positive integers. Describe a simple relation of $m$ and $n$ which is equivalent to $\mathbb{Q}[\zeta_n] = \mathbb{Q}[\zeta_m]$

A student solving exercises on Galois theory made the following argument. Consider $\zeta_{19}$. She could see that $-\zeta_{19}$ is one of primitive $38$th root of unity. So she conclude $\mathbb{Q}[\zeta_{19}] = \mathbb{Q}[\zeta_{38}]$. That was right. But it seems that she thought those two fields are not same. I have told her the fact. However I cannot tell a simple criterion amounts to the above question.
Would you please help me?
 A: Assuming that $n \ge m$ WLOG, $\mathbb{Q}[\zeta_n] = \mathbb{Q}[\zeta_m]$ (note that $=$ is unambiguous here only because these extensions are Galois) iff either $n = m$ or $m$ is odd and $n = 2m$.
Probably there are a couple of different ways to do this but here's a relatively straightforward one. $\mathbb{Q}[\zeta_n]$ and $\mathbb{Q}[\zeta_m]$ both embed into $K = \mathbb{Q}[\zeta_{\ell}]$ where $\ell = \text{lcm}(n, m)$, and as subfields of $K$ they are determined by the Galois correspondence: they are the fixed fields of the subgroups of the Galois group $G = U(\ell)$ given by the kernels of the quotient maps $U(\ell) \to U(n), U(\ell) \to U(m)$, and $\mathbb{Q}[\zeta_n] = \mathbb{Q}[\zeta_m]$ (again, this is only unambiguous because they're Galois) iff these subgroups are equal.
By the Chinese remainder theorem, to understand these kernels it suffices to work one prime at a time and understand the kernels of the quotient maps $U(p^a) \to U(p^b), a \ge b$. If $p$ is odd this is a surjective homomorphism between cyclic groups so its kernel is the unique subgroup of $U(p^a)$ of index $p^{b-a}$ and in particular, for fixed $a$ its size uniquely determines the value of $b$. It follows that $\nu_p(n) = \nu_p(m)$ for all odd primes $p$.
So the only possible discrepancy between $n$ and $m$ is at the prime $2$. At this point we don't have to keep analyzing the group of units and we can argue as follows: the condition $\varphi(n) = \varphi(m)$ implies, after dividing out the contribution of the odd primes, that either
$$2^{\nu_2(n) - 1} = 2^{\nu_2(m) - 1}$$
(and hence that $n = m$) if $\nu_2(n), \nu_2(m) \ge 1$, or that $\nu_2(m) = 0$ and $\nu_2(n) = 0, 1$. So either $n = m$, or $m$ is odd and $n = 2m$, as stated. In this second case we have that $- \zeta_m$ is a primitive $n^{th}$ root of unity, generalizing your example.
