Determine the Galois group of $x^{15} - 1 $ over $\mathbb{Q}(i)$ and all its intermediate field

Question

I know already that the Galois group of $x^{15}-1$ over $\mathbb Q$ should be the units of $\mathbb Z_{15}$ i.e. $1, 2, 4, 7, 8, 11, 13, 14$. It is commutative, so can only be either $\mathbb Z_2 \times\mathbb Z_4$ or $\mathbb Z_2 \times\mathbb Z_2 \times\mathbb Z_2$. Furthermore, $7, 13$ are both of order $4$ so it can only be $\mathbb Z_2 \times\mathbb Z_4$. Now the group corresponding to $\mathbb Q(i)$ should be of index $2$, thus of order $4$. Then it is either $\mathbb Z_2 \times\mathbb Z_2$ or $\mathbb Z_4$. How to continue this line of reasoning to get its Galois group and all intermediate fields?

Answer 1

I would attack the question in a rather different manner.

Please note that $i\notin\Bbb Q(\zeta_{15})$. A simple (but high-powered) argument for seeing this is the observation that if $n$ is odd, the primes ramifying in $\Bbb Q(\zeta_n)$ are exactly those dividing $n$. But $\Bbb Q(i)$ is ramified at $2$.

If you believe the above claim, you see that the Theorem on Natural Irrationalities applies, according to which (in this case) $\text{Gal}^{\Bbb Q(\zeta_{15})}_{\Bbb Q}\cong\text{Gal}^{\Bbb Q(\zeta_{15},i)}_{\Bbb Q(i)}$.

Consequently, we need only consider the extension $\Bbb Q(\zeta_{15})\supset\Bbb Q$ and its intermediate fields. The intermediate fields of the translated extension $\Bbb Q(\zeta_{15},i)\supset\Bbb Q(i)$ correspond to those in the simpler situation, by adjoining $i$ to each of them.

Again, since $\Bbb Q(\zeta_5)\cap\Bbb Q(\zeta_3)=\Bbb Q$ and $\Bbb Q(\zeta_{15})=\Bbb Q(\zeta_5)\Bbb Q(\zeta_3)$, the Galois group is the product of the two Galois groups over $\Bbb Q$, namely $C_4\oplus C_2$, where I’m using the notation $C_m$ for a cyclic group of order $m$.

I won’t work through the list of subgroups of $C_4\oplus C_2$, but once you’ve identified the (unique) field between $\Bbb Q$ and $\Bbb Q(\zeta_5)$, the job is relatively easy. In any event, I leave that to you.