Cyclotomic polynomials, properties. Let $F$ be a field of characteristic prime to $n$, and let $F^a$ be an algebraic closure of $F$. Let $\zeta$ be a primitive $n$th root of unity in $F^a$. I know that the monic polynomial $\Phi_n(X)$ factorizes as $$\Phi_n(X) = \prod_{m \in (\mathbb{Z}/n)^\times} (X - \zeta^m).$$I also know that there is a canonical injective group homomorphism $\chi: \text{Gal}(F(\zeta)/F) \to (\mathbb{Z}/n)^\times$ that is characterized by the property that $g(\alpha) = \alpha^{\chi(g)}$ for any $g \in \text{Gal}(F(\zeta)/F)$ and any $\alpha \in \mu_n$.
What is the easiest and quickest way see that the following properties are true/where can I find a reference to their proofs?

Let $S \subset (\mathbb{Z}/n)^\times$ be the image of the above homomorphism $\chi$.

*

*If $m_1$ and $m_2$ are two elements of $(\mathbb{Z}/n)^\times$ then $\zeta^{m_1}$ and $\zeta^{m_2}$ are conjugate to another if and only if $m_1$ and $m_2$ lie in the same coset of $S$ in $(\mathbb{Z}/n)^\times$.


*Let $S = S_1, \dots, S_s$ denote the cosets of $S$ in $(\mathbb{Z}/n)^\times$. Then for each $i = 1, \dots, s$, the polynomial $f_i = \prod_{m \in S_i} (X - \zeta^m)$ is an irreducible element of $F[X]$.


*$\Phi_n(X) = f_1(X) \dots f_s(X)$.

Much thanks in advance.
 A: *

*$\zeta^{m_1}$ and $\zeta^{m_2}$ are conjugate if and only if there exists $g \in \text{Gal}(F(\zeta)/F)$ such that $g(\zeta^{m_1}) = \zeta^{m_2}$. This is the case if and only if there exists $g \in \text{Gal}(F(\zeta)/F)$ such that $\chi(g)m_1 = m_2$, which happens if and only if $m_1$ and $m_2$ lie in the same coset of $\text{Im}\,\chi = S$ in $(\mathbb{Z}/n\mathbb{Z})^\times$.

*We choose coset representatives $t_i$ so that $S_i = t_iS$; we have that $(\mathbb{Z}/n\mathbb{Z})^\times$ is commutative, so we need not worry about left or right. By (1),$$f_i(X) = \prod_{g \in \text{Gal}(F(\zeta)/F)} (X - \zeta^{t_i \chi(g)}) = \prod_{g \in \text{Gal}(F(\zeta)/F)} (X - g(\zeta^{t_i})).$$As this expands, it is clear that all coefficients are stable under the action of $\text{Gal}(F(\zeta)/F)$ and so $f_i \in F(\zeta)^G[X] = F[X]$, as $F(\zeta)/F$ is Galois. Moreover, it is the minimal polynomial of $\zeta^{t_i}$ over $F$ (by (1)), and so is irreducible.

*Each element of $(\mathbb{Z}/n\mathbb{Z})^\times$ appears in exactly one coset of $S$, and thus, appears as a power of $\zeta$ in exact one of the $f_i$. The equality in question is now clear.

A: I am really not sure of this, but here's what I could do!
Let $\tau$ be in the Galois group. $\tau$ is uniquely determined by what it does to $\zeta$. And $\tau_{k}(\zeta)=\zeta^{k}$. Now if $m_{1}$ and $m_{2}$ lie in the same coset of $S$, then $m_{1}m_{2}^{-1} \in S$, So $m_{1}m_{2}^{-1}=t$ for some $t$ in $S$. But we know that we can find an elememt $\tau_{t}$ in the Galois group such that $\tau_{t}(\zeta)=\zeta^{t}$, so $\tau_{t}(\zeta^{m_{2}})=\zeta^{m_{2}t}=\zeta^{m_{1}}$.
Assume they're conjugate, then we can find $\sigma$ in the Galois group such that $\sigma(\zeta^{m_{1}})=\zeta^{m_{2}}$, and hence we can find $t \in S$ such that $\sigma(\zeta)=\zeta^{k}$ so $\zeta^{km_{1}}=\zeta^{m_{2}}$ and since $\zeta$ is primitive, $m_{1}k=m_{2}$ so $m_{2}m_{1}^{-1}=k \in S$.
For the second, your polynomial indeed belongs to $F[X]$ as it's invariant under the action of the Galois group, by using what we proved in part 1. Now, if we have a proper irreducible factor $g_{i}(x)$ of $f_{i}(x)$, then $g_{i}(x)=\prod\limits_{k \in T_{i}}(x-\zeta^{k})$, where $T_{i}$ is a proper subset of $S_{i}$. But, also using the first part, you can prove that this polynomial does not belong to $F[X]$ as you can produce an automorphism that does not fix it!.
Now for third part, I think you can use the fact that the cosets of $S$ partition $(\mathbb{Z}/n\mathbb{Z})^{\star}$
