Computing Order of Galois Groups

Question

I would like to compute the order of Galois Group of following polynomials over $\mathbb{Q}$. I am stuck here so I need some help

1. $X^4-1$
2. $X^4 -3$
3. $X^6+3$
4. $X^6-3$
5. $X^6+4$
6. $X^8+1$
7. $X^8+2$
8. $X^8-2$

Since we are talking about splitting fields over a field of characteristic $0$, they are Galois Extensions. Hence the order of the Galois Group is equal to the degree of the extension.

For the first two examples, It was very easy to determine the splitting fields but it gets complicated for me after that. I would appreciate if someone can shed some light over $(c)-(g)$. I really need this to proceed further with Galois Theory.

Answer 1

Some hints:

$4.$ The roots are $\alpha\omega^k$, where $\alpha=\sqrt[6]{3}$ and $\omega$ is a sixth-root of unity. So the splitting field is $\mathbb{Q}(\alpha, \omega) $. Note that $\omega=1/2+i\sqrt3/2$. So $\mathbb{Q}(\alpha, \omega)= \mathbb{Q}(\alpha, \sqrt3 i)$. But $\sqrt3=\alpha^3\in \mathbb{Q}(\alpha, \sqrt3 i)$, so the splitting field simplifies to $\mathbb{Q}(\alpha,i)$. Since one generator is complex and the other is real, is easy to compute that the degree of extension is $12$. A group of order $12$ generated by a element of order $6$ and another of order $2$... It seems like dihedral group. Try proving it finding two automorphisms that give the structure to the group.

$6.$ Note that $x^8+1$ is the cyclotomic polynomial of order $16$ so it's Galois group is directly derived from here.

The cases $3,5,6,7$ are very similar to the $4$.