TITLE: Galois group of $x^4-2$
QUESTION [22 upvotes]: I am trying to explicitly compute the Galois group of $x^4-2$ over $\mathbb{Q}$. I found that the resolvent polynomial is reducible and the order of the Galois group is $8$ using the splitting field $K=\mathbb{Q}(2^{1/4}, i)$. Hence I need to find 8 automorphisms. Thus do I just map elements of the same order to each other that fix the base field $\mathbb{Q}$ $2^{1/4}$ to $i2^{1/4}$

REPLY [24 votes]: You can see that $i^n\sqrt[4]{2},\, 0\le n\le 3$ are all the roots, so that
$$K=\Bbb Q(i,\sqrt[4]{2})$$
is the splitting field for the polynomial. Since $L=\Bbb Q(\sqrt[4]{2})$ is real of degree $4$, we see that $K$ is a proper extension of $L$, and since $[\Bbb Q(i):\Bbb Q]=2$ we see the total degree of the extension is $2\cdot 4=8$. But then we have that $\operatorname{Gal}\left(K/\Bbb Q\right)\le S_4$ is a subgroup of $S_4$ of order $8$. This implies it is a Sylow-$2$ subgroup of $S_4$, all of which are isomorphic--by the second Sylow theorem. We know that $D_8$, the dihedral group of order $8$, is such a subgroup, so that gives the isomorphism type.
But then you know what to look for as explicit representations go, you note that relative to the ordering
$$\alpha_j=i^j\sqrt[4]{2}, 1\le j\le 4$$
we have the $4$-cycle $(1234)$ given by the automorphism
$$\begin{cases}\sqrt[4]{2}\mapsto i\sqrt[4]{2} \\ i\mapsto i\end{cases}$$
which is enough to totally determine it, since those are generators of the extension. Clearly also
$$\begin{cases}\sqrt[4]{2}\mapsto \sqrt[4]{2} \\ i\mapsto -i\end{cases}$$
is represented by the transposition $(13)$, and these two generate the group, so give you everything you need for a fully explicit description.