TITLE: Galois group of $x^4-5$
QUESTION [7 upvotes]: How do I find the Galois group of $x^4-5$ over $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt5)$ and $\mathbb{Q}(\sqrt{-5})$?
I've managed to do so over $\mathbb{Q}$ but I don't know how to find the others.
I'd appreciate any help.
Thanks!

REPLY [25 votes]: You have already found that the splitting field is
$E=\mathbb{Q}(\mathrm{i},\sqrt[4]{5})$
and that $\operatorname{Gal}(E/\mathbb{Q})\cong D_8$ (I use the algebraic
convention where the index denotes the group order, not the vertex count).
Therefore the other fields you cited are intermediate fields $K$
between $E$ and $\mathbb{Q}$.
Use the fundamental theorem of Galois theory,
that is, find the subgroup of $\operatorname{Gal}(E/\mathbb{Q})$
whose members fix $K$.
First, let us find a suitable presentation for
$G = \operatorname{Gal}(E/\mathbb{Q})$.
A straightforward choice is $G = \langle\tau, \sigma\rangle$
where $\tau$ is complex conjugation, thus
fixing $\sqrt[4]{5}$ and sending $\mathrm{i}$ to $\tau(\mathrm{i})=-\mathrm{i}$,
and $\sigma$ fixes $\mathrm{i}$ while rotating $\sqrt[4]{5}$
to $\sigma(\sqrt[4]{5})=\mathrm{i}\sqrt[4]{5}$.
Therefore $\tau$ has order $2$, $\sigma$ has order $4$,
and $\sigma\tau=\tau\sigma^{-1}$.
It remains to find subgroups of $G$ that fix the various intermediate fields
$K$. The index of such a subgroup in $G$ must equal the extension degree
of its fixed field over $\mathbb{Q}$.
That is, subgroups of order $d$ dividing $8$
correspond to intermediate fields of degree $8/d$ over $\mathbb{Q}$.
For example, $\sigma^2$ fixes $\mathrm{i}$ and $\sqrt{5}$, therefore
the order-$2$ subgroup $\langle\sigma^2\rangle$ fixes the degree-$4$
extension field $\mathbb{Q}(\mathrm{i},\sqrt{5})$.
It fixes no larger field, as that would violate the index condition.
Likewise, $\tau$ fixes $\sqrt[4]{5}$ and thus fixes $\sqrt{5}$ as well,
so the order-$4$ subgroup $\langle\tau,\sigma^2\rangle$
fixes the degree-$2$ extension field $\mathbb{Q}(\sqrt{5})$.
I have made a diagram that contains some more intermediate fields.
If I got it right (check it!), we have the following correspondence,
showing fields together with the subgroups of $G$ that fix them.
Arrows mean "extends" for fields and "is subgroup of" for Galois groups.
$$\begin{matrix}
 &&&& \substack{\mathbb{Q}(\mathrm{i},\sqrt[4]{5})\\\{1\}}
\\ &&& \swarrow & \downarrow & \searrow
\\ && \substack{\mathbb{Q}(\mathrm{i},\sqrt{5})
 \\\langle\sigma^2\rangle}
 && \substack{\mathbb{Q}(\sqrt[4]{5})
 \\\langle\tau\rangle}
 && \substack{\mathbb{Q}(\mathrm{i}\sqrt[4]{5})
 \\\langle\tau\sigma^2\rangle}
\\ & \swarrow & \downarrow & \searrow & \downarrow & \swarrow
\\ \substack{\mathbb{Q}(\mathrm{i})
 \\\langle\sigma\rangle\cong C_4}
 && \substack{\mathbb{Q}(\mathrm{i}\sqrt{5})
 \\\langle\tau\sigma,\sigma^2\rangle\cong V_4}
 && \substack{\mathbb{Q}(\sqrt{5})
 \\\langle\tau,\sigma^2\rangle\cong V_4}
\\ & \searrow & \downarrow & \swarrow
\\ && \substack{\mathbb{Q}
 \\\langle\tau,\sigma\rangle\cong D_8}
\end{matrix}$$
The third row contains what your question has sought.
$C_4$ means a cyclic group of order $4$, $V_4$ means Klein's Four group.
Note that the diagram is not complete: I left out e.g.
$\mathbb{Q}\left((1-\mathrm{i})\sqrt[4]{5}\right)$
which is fixed by $\tau\sigma$ and extends $\mathbb{Q}(\mathrm{i}\sqrt{5})$.
You may want to complete the diagram to get the full picture.