I have been reading a paper which uses without proof the following result:
Let $n \in \mathbb N$ and $p$ be a prime divisor of $n$. For a positive integer $m$, by $\zeta_m$ we mean a primitive $m$-th root of unity, so that $\mathbb Q(\zeta_m)$ is just the $m$-th cyclotomic field. Then the Galois group $G:= \text{Gal}(\mathbb Q(\zeta_{np}) / \mathbb Q(\zeta_{n/p}))$ is cyclic.
It is not clear to me why this holds. I did notice that the order of $G$ is equal to $p(p-1)$ (respectively $p^2$) when $p$ exactly divides $n$ (respectively when $p^2|n$), but there is at least one non-cyclic group of these orders. I also tried using the result that the Galois group of a compositum of Galois extensions is the product of the Galois groups of the concerned extensions, but to no avail. What a I missing? Thanks.
Best Answer
I'll give details for the case when $p$ divides $n$ but $p^2$ doesn't divide $n$ and leave you to generalise.
Then $n=pm$ for some integer $m$ coprime to $p$ and we have $\operatorname{Gal}(\mathbb{Q}(\zeta_{n/p})/\mathbb{Q}) \cong (\mathbb{Z}/m\mathbb{Z})^{\times}$ and similarly $\operatorname{Gal}(\mathbb{Q}(\zeta_{np})/\mathbb{Q}) \cong (\mathbb{Z}/p^2m\mathbb{Z})^{\times}$.
By the fundamental theorem of Galois theory, $\operatorname{Gal}(\mathbb{Q}(\zeta_{np})/\mathbb{Q})/ \operatorname{Gal}(\mathbb{Q}(\zeta_{np})/\mathbb{Q}(\zeta_{n/p})) \cong \operatorname{Gal}(\mathbb{Q}(\zeta_{n/p})/\mathbb{Q})$.
Now note that since $p \nmid m$, $(\mathbb{Z}/p^2m\mathbb{Z})^{\times} \cong (\mathbb{Z}/p^2\mathbb{Z})^{\times} \times (\mathbb{Z}/m\mathbb{Z})^{\times}$ (by the Chinese Remainder Theorem). Now we have (abstractly) $(\mathbb{Z}/p^2m\mathbb{Z})^{\times} / \operatorname{Gal}(\mathbb{Q}(\zeta_{np})/\mathbb{Q}(\zeta_{n/p})) \cong (\mathbb{Z}/m\mathbb{Z})^{\times}$ and so by the above isomorphism $\operatorname{Gal}(\mathbb{Q}(\zeta_{np})/\mathbb{Q}(\zeta_{n/p})) \cong (\mathbb{Z}/p^2\mathbb{Z})^{\times} $, which is well known to be cyclic of order $p(p-1)$.