# Galois group of $x^4-2$

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 $i2^{1/4}$

You can see that $i^n\sqrt{2},\, 0\le n\le 3$ are all the roots, so that

$$K=\Bbb Q(i,\sqrt{2})$$

is the splitting field for the polynomial. Since $L=\Bbb Q(\sqrt{2})$ 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

$$\alpha_j=i^j\sqrt{2}, 1\le j\le 4$$

we have the $4$-cycle $(1234)$ given by the automorphism

$$\begin{cases}\sqrt{2}\mapsto i\sqrt{2} \\ i\mapsto i\end{cases}$$

which is enough to totally determine it, since those are generators of the extension. Clearly also

$$\begin{cases}\sqrt{2}\mapsto \sqrt{2} \\ i\mapsto -i\end{cases}$$

is represented by the transposition $(13)$, and these two generate the group, so give you everything you need for a fully explicit description.

• I believe the OP is not asking for the isomorphism type, but rather for the actual elements of the group. Apr 12 '15 at 21:28
• @MattSamuel you may be right, I've edited it to reflect this. Thanks for the comment! Apr 12 '15 at 21:35
• @AdamHughes: Why do we need to specify where $i$ goes? Wouldn't it be sufficient to know just where the roots go: Say for instance I have $x_1=\sqrt{2}$, $x_2=-\sqrt{2}$, $x_3=i\sqrt{2}$, $x_4=-i\sqrt{2}$. I can choose to map $x_1 \to x_2$ and then $x_1\to x_2 \to x_3 \to x_4$ and get a $2$ and $4-$ cycle respectively. Apr 15 '15 at 6:18
• @RutherfordMark that's fine, you're doing the same thing, only with different generators for the field, I chose $i,\sqrt{2}$ because those were the field generators you gave in the original problem. Clearly if you know one you know the other, since $\sigma(i\sqrt{2})=\sigma(i)\sigma(\sqrt{2})$. Apr 15 '15 at 7:42
• @Adam Hughes If we want to get the subgroups of order 4 of the Galois group $G(K/ \mathbb Q)$ how can we do this? I mean if we want to get the subgroups of order $2$ we know that they are of the form $\{e,\sigma\}$ such that $\sigma^2=e$ where $e$ is the nuetral element. So what about the subgroups of order 4 for example? Jan 30 '19 at 21:19

You should note that the Galois group is isomorphic to a subgroup of $S_4$, since any element of the Galois group will be determined by how it permutes the four roots of $x^4-2$. The order 8 subgroups of $S_4$ are isomorphic to the dihedral group with 8 elements.

If you want to know exactly what the eight maps are, there are four choices as to where $\sqrt{2}$ is sent under some element of the Galois group. That tells you where the root $-\sqrt{2}$ is sent as well, and then you have two choices as to where $i\sqrt{2}$ is sent.