# Question about unique subfields in cyclotomic fields.

abstract-algebraalgebraic-number-theorynumber theory

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:

1. 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., $$F_{2}=\mathbb{Q}\left(\sqrt{\left(-1\right)^{\frac{p-1}{2}}}p\right)$$.
2. $$F_{d_{1}}\subseteq F_{d_{2}}$$ if and only if $$d_{1}\mid d_{2}$$.
3. 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 $$F_{d}$$.
4. (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!

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 $$p^*= 13$$.