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

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.

• Sorry, I am confused, aren't we supposed to compute $Gal_Q^{Q(\zeta_{15})}$? – penny Jun 3 at 0:55
• 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 $\Bbb Q(i)$. – Lubin Jun 3 at 1:11
• 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? – penny Jun 3 at 1:22
• 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. – Lubin Jun 3 at 3:30