Find the splitting field of $x^{12}-1$
We can write the roots as $$x = e^{in\frac{2\pi}{12}}, x = 1,\cdots, 12$$
Since all the roots can be achivied through self multiplication of $e^{i\frac{2\pi}{12}}$, the splitting field is $\mathbb{Q}(e^{i\frac{2\pi}{12}}) = \mathbb{Q}(\sqrt{3}/2 + i/2):\mathbb{Q}$. How to take $[\mathbb{Q}(e^{i\frac{2\pi}{12}}):\mathbb{Q}]$?
$$[\mathbb{Q}(\sqrt{3}/2 + i/2):\mathbb{Q}] = [\mathbb{Q}(\sqrt{3} + i):\mathbb{Q}]$$
I thought about finding the basis of $\mathbb{Q}(\sqrt{3}+i)$ over $\mathbb{Q}$. $\mathbb{Q}(\sqrt{3}+i)$ is basically $\mathbb{Q}$ adjoined with every multiple of $\sqrt{3}+i$ plus every multiple of its inverse, that is, multiples of $\sqrt{3}-i$. We're at least talking about the set
$$\{a + b(\sqrt{3}+i) + c(\sqrt{3}-i)\}$$
BUT I think that
$$\{a + b(\sqrt{3}+i)\}$$
is already enough because we can get $\sqrt{3}-i$ from this set. So I think the splitting field should have degree $2$. Am I right? In this case, the Galois of this splitting field would be made by the identity automorphism and conjugate automorphism.
We can write the roots as $\zeta^k$ where $0 \le k < 12$.
Then, let $r$ be an automorphism of $\Bbb Q(\zeta)/\Bbb Q$ that takes $\zeta$ to $\zeta^a$. We know that $r(\zeta^k) = r(\zeta)^k = \zeta^{ak}$.
Therefore, automorphisms of $\Bbb Q(\zeta)/\Bbb Q$ are automorphisms of $\Bbb Z_{12}$, which is $(\Bbb Z_{12})^* \cong V_4$.
The minimal polynomial of $\zeta$ is $\Phi_{12} = x^4-x^2+1$ (cyclotomic polynomial).
$x^{12}-1 = (x-1)(x+1)(x^2+x+1)(x^2-x+1)(x^4-x^2+1)$