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?


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.

  • $\begingroup$ Sorry, I am confused, aren't we supposed to compute $Gal_Q^{Q(\zeta_{15})}$? $\endgroup$ – penny Jun 3 at 0:55
  • $\begingroup$ No, you said, “over $\Bbb Q(i)$”, but $\Bbb Q(\zeta_{15})$ doesn’t contain $i$, so that you have to talk about the field gotten by adjoining the roots of $X^{15}-1$ to $\Bbb Q(i)$. $\endgroup$ – Lubin Jun 3 at 1:11
  • $\begingroup$ I see, but that is strange since the question is phrased to make me believe that $i$ is in there. Would you please explain what is the meaning of ramifying? $\endgroup$ – penny Jun 3 at 1:22
  • $\begingroup$ Yeah, ramification has to do with what happens to a prime $p\in\Bbb Z$ when you look at the ring of integers $\mathcal O_K$ of an extension $K\supset\Bbb Q$. If $p\mathcal O_K$ is a product of distinct $K$-primes, then one says that $p$ is unramified in $K$, and if there’s a repetition of a factor when you express $\mathcal O_K$ as a product of prime ideals, you say that $p$ ramifies in $K$. As I say, it’s rather advanced. You should be able to show that $i\notin\Bbb Q(\zeta_{15})$ directly. $\endgroup$ – Lubin Jun 3 at 3:30

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