Structure of Galoisgroup

Question

Let $K=\mathbb{Q}(\zeta_3,i,\sqrt{3})$, whereas $\zeta_3=\exp\big(\frac{2\pi i}{3}\big)$. How can one calculate the structure of the Galoisgroup?

Answer 1

First note $(2\zeta_3+1)^2 = 4\zeta_3^2+4\zeta_3+1 = 4(-\zeta_3-1) + 4\zeta_3+1 = -3$, so $\mathbb{Q}(\zeta_3) = \mathbb{Q}(i\sqrt 3)$, and we see that $K = \mathbb{Q}(i,\sqrt 3)$.

Now $a+bi$ cannot be a square root of 3 as $(a+bi)^2 = a^2-b^2+2abi$, so we would have $2ab = 0$ so either $a$ or $b$ is $0$, but 3 is not of the form $a^2$ or $-b^2$ for any rational $a,b$.

Thus $K$ is a degree 4 extension (i.e. $\sqrt 3 \notin \mathbb{Q}(\sqrt 3)$), and is the compositum of the Galois subextensions $\mathbb{Q}(i),\mathbb{Q}(\sqrt 3)$, thus is Galois.

We can see its Galois group is $C_2\times C_2$ in a few different ways.

One way is to see that since $\mathbb{Q}(i),\mathbb{Q}(\sqrt 3)$ are linearly disjoint (meaning their intersection is $\mathbb{Q}$), the injective natural map $Gal(K/\mathbb{Q}) \to Gal(\mathbb{Q}(\sqrt 3/\mathbb{Q}) \times Gal(\mathbb{Q}(i)/\mathbb{Q})$ given by restriction must also be surjective as both groups are order $4$, so it is an isomorphism.

Another way is to note that two degree two sub-extensions exist, so the Galois group must have two subgroups of order $2$, but there is only one such group of order $4$.

Alternatively note that $K = \mathbb{Q}(\zeta_4,\zeta_3) = \mathbb{Q}(\zeta_{12})$, for which the Galois group is naturally isomorphic to $\mathbb{Z}/12\mathbb{Z}^\times$, where the isomorphism is given by $a \mapsto \tau_a$, where $\tau_a$ is the automorphism sending $\zeta_{12}$ to $\zeta_{12}^a$. It is easy to see this map is injective since if $\zeta_{12} = \zeta_{12}^a$, $\zeta_{12}$ is an $a-1^{th}$ root of unity, but then $a$ must be congruent to $1$ mod 12 by definition of $\zeta_{12}$. Thus it is an isomorphism again since both groups are order 4.