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!

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.

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

$$[\Bbb Q(\sqrt[4]5\,,\,i):\Bbb Q(\sqrt5 i]=4$$