Question about unique subfields in cyclotomic fields. I'm self-studying Number Theory and searching for exercises for number theory I came across this assignment which has this exercise:
(Special Cyclotomic Case). Let $p$ be an odd prime, $\omega=\omega_{p}$
a primitive $p$th root of unity. Show all of the following:

*

*If $d\mid p-1$, then there exists a unique subfield $F_{d}$ of $\mathbb{Q}\left(\omega\right)$
of degree $d$ over $\mathbb{Q}$, e.g., $F_{2}=\mathbb{Q}\left(\sqrt{\left(-1\right)^{\frac{p-1}{2}}}p\right)$.

*$F_{d_{1}}\subseteq F_{d_{2}}$ if and only if $d_{1}\mid d_{2}$.

*If $q\neq p$ is an odd prime and $d\mid p-1$, then $q$ is a $d$th
power modulo $p$ if and only if $q$ splits completely in $F_{d}$.

*(Quadratic Reciprocity) If $p$ and $q$ are two odd primes, then
the Legendre symbol $\left(\dfrac{p}{q}\right)=1$ if and only if
$q$ splits completely in $F_{2}$ if and only if the Legendre Symbol
$\left(\dfrac{d_{\mathbb{Q}\left(\omega\right)}}{q}\right)=1$. In
particular,
$$\left(\dfrac{q}{p}\right)\left(\dfrac{p}{q}\right)=\left(-1\right)^{\frac{p-1}{2}\frac{q-1}{2}}.$$
I'm struggling currently to prove the second question which is needed to show the third one (I suppose). I have no clue how to relate the degree with the unique subfields of $\mathbb{Q}\left(\omega\right)$. Any ideas? Thanks in advance!
 A: According to Galois theory, the sub-field structure of $\mathbb{Q}(\omega_p)$ (where $\omega_p = e^{\frac{2\pi i}{p}}$) corresponds with the subgroup structure of $Aut(\mathbb{Q}(\omega_p))$, the automorphism group of $\mathbb{Q}(\omega_p)$.


Clearly, when $\sigma \in Aut(\mathbb{Q}(\omega_p))$, then $\left(\sigma(\omega_p)\right)^p = \sigma(\omega_p^p) = \sigma(1) = 1$, so $\sigma(\omega_p)$ is again a $p-th$ order root of unity, hence we must have $\sigma(\omega_p) = \omega_p^j$ for some $j \in \{1,2,...,p-1\}$.
Also, defining $\sigma_j$ by $\sigma_j(\omega_p) = \omega_p^j$, we see that $\sigma_j\left(\sigma_k(\omega_p)\right) = \sigma_j(\omega_p^k) = \omega_p^{jk}$, implying that $Aut(\mathbb{Q}(\omega_p))$ is isomorphic to $\mathbb{F}_p^*$, so it is cyclic, generated by $\sigma_g$ where $g$ is a generator of $\mathbb{F}_p^*$.


The subgroups of a cyclic group $<\negthickspace \sigma \negthickspace>$ are $<\negthickspace\sigma^e\negthickspace>$ where $e$ divides the group order, $p-1$ in this case. Thus, for example, for $p=13$ we get the following correspondence between subfields of $\mathbb{Q}(\omega_p)$ and subgroups of $\mathbb{F}_p^*$:

Where $p^*=\left(\frac{-1}{p}\right) p$, in the example $p^*= 13$.
This answers (1) and (2).
I leave (3) and (4) for you :-).
