# If $K\cap\Bbb Q^{\text{cycl}}=\Bbb Q(\zeta_m)$ and $K/\Bbb Q$ Galois, then $\text{Gal}(K(\zeta_n)/K)\cong\text{Gal}(\Bbb Q(\zeta_n)/\Bbb Q(\zeta_m))$

$\DeclareMathOperator{\Gal}{Gal}$ Here is my argument: Induction on the number of primes dividing $n/m$.

If there are two primes (i.e., $K(\zeta_n) = K(\zeta_{q_1},\zeta_{q_2})$, where $q_i=p_i^{e_i}$), then we know:

$$\Gal(K(\zeta_{q_1},\zeta_{q_2})/K) \cong \\ \cong \{(\sigma,\tau)\in \Gal(K(\zeta_{q_1})/K)\times \Gal(K(\zeta_{q_2})/K) \;|\;\sigma_{|K(\zeta_{q_1})\cap K(\zeta_{q_2})} = \tau_{|K(\zeta_{q_1})\cap K(\zeta_{q_2})}\}$$

Since $K(\zeta_{q_1})\cap K(\zeta_{q_2})=K$, then $\Gal(K(\zeta_{q_1},\zeta_{q_2})/K) \cong \Gal(K(\zeta_{q_1})/K) \times \Gal(K(\zeta_{q_2})/K)$.

Since $K\cap\Bbb Q^{\text{cycl}}=\Bbb Q(\zeta_m)$ then $K(\zeta_{q_i})\cap\Bbb Q^{\text{cycl}}=\Bbb Q(\zeta_m, \zeta_{q_i})$, and taking Galois groups, the result follows.

The induction step is similar.

However, I didn't use the fact that $K/\Bbb Q$ is Galois anywhere. What am I missing?

• @Timon OK, I don't mind your edit, but why does the closing bracket appear on its own line after your edit? I liked my way better... – Alex Apr 14 '16 at 17:28
• I found the answer. It actually requires $K/\Bbb Q(\zeta_m)$ to be Galois. If anyone needs me to, I can post it. – Alex Apr 14 '16 at 18:46
• I think $\times$ is better than 'x'; $\operatorname{Gal}(\dots)$ is better than 'Gal($\dots$)'; $\mathbb{Q}^{\text{cycl}}$ is better than $\Bbb Q^{cycl}$; consistency when switching math/text mode is better than inconsistency. I think that expression cannot be both readable and in one line. However, make your post as you prefer ;) – Τίμων Apr 14 '16 at 19:37
• @Timon Once again, I don't mind the edit, I can deal with it one way or another, but why are there now backslashes in front of the Gal in the title? I don't mind having it in different font or whatever, but I want it to be easily readable. – Alex Apr 14 '16 at 19:48
• @Alex since it is your question, it is your choice how to write it. You should change anything you don't like. If you have an answer feel free to post it! – Mathmo123 Apr 15 '16 at 0:22