Galois group of $x^4-5$ 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!
 A: Since $\;x^4-5=(x^2-\sqrt5)(x^2+\sqrt5)\;$ , the polynomial's roots are $\;\pm\sqrt[4]5\;,\;\pm\sqrt[4]5\,i\;$ .
But $\;x^4-5\in\Bbb Q(i)[x]\;$ is irreducible, otherwise either
$$\sqrt[4]5\in \Bbb Q(i)\implies \exists\,a,b\in\Bbb Q\;\;s.t.\;\;\sqrt[4]5=a+bi$$
which we know is possible in $\;\Bbb C\;$ iff $\;a=\sqrt[4]5\;,\;\;b=0\;$ (equality in $\;\Bbb C\;$) , or else it can be factored in the product of two quadratics over $\;\Bbb Q(i)\;$ , which is now easy to see it's impossible (as the factoring above happens in $\;\Bbb C[x]\;$ , which contains the involved fields's polynomial rings)
Thus, 
$$[\Bbb Q(i,\sqrt[4]5):\Bbb Q(i)]=4\implies \text{the Galois group wanted is the cyclic}\;\;C_4\;\;or\;\;C_2\times C_2$$  Knowing what the full Galois group is shall help to find out which one it is .
Try now to do something similar for the other cases.
Added on request: For example, over $\;K:=\Bbb Q(\sqrt{-5})=\Bbb Q(\sqrt5\,i)\;$:
The polynomial $\;x^4-5\in K[x]\;$ is irreducible, otherwise either
$$\pm\sqrt5\in K\iff \exists\,a,b\in\Bbb Q\;\;s.t.\;\;\pm\sqrt5=a+b\sqrt5i$$
or else
$$\pm\sqrt[4]5\,,\,\pm\sqrt[4]5i\in K\implies \exists\,a,b\in\Bbb Q\;\;s.t.\;\;\begin{cases}\pm\sqrt[4]5\\\pm\sqrt[4]5i\end{cases}=a+b\sqrt5i\implies$$
and both cases lead us to contradiction (check this!), thus
$$[\Bbb Q(\sqrt[4]5\,,\,i):\Bbb Q(\sqrt5 i]=4$$
A: 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.
