Galois group of $x^6+1$

Question

$x^6+1$ has $6$ roots: $i,i\xi,i\xi^2,i\xi^3,i\xi^4,i\xi^5$ where $\xi=e^{\tfrac{2\pi i}{6}}$. Since $x^{12}-1=(x^6-1)(x^6+1)$ the splitting field of $x^{12}-1$ contains the splitting field of $x^6+1$ and therefore the degree of the splitting field of $x^6+1$ must be $4$ or $2$. Can anyone help me out here whether the degree is 4 or 2?

Answer 1

The splitting field of $ x^{12}-1 $ is the same as the splitting field of $x^6+1$. Actually, it is the $12$-th cyclotomic field and its degree is $\varphi(12)=4$.

The minimal polynomial is easy to find with simple high school formulae: $$x^6+1=(x^2)^3+1=(x^2+1)(x^4-x^2+1)$$ A generator of the splitting field is $\,\mathrm e^{\tfrac{\mathrm i\pi}{6}}$ thus has minimal polynomial: $\, x^4-x^2+1$. (We check its factorisation in $\mathrm R[x]$ is $(x^2-\sqrt3x+1)(x^2+\sqrt3x+1)$, hence it is irreducible over $\mathrm Q$).