Galois Group of $x^{4}+7$ I am trying to find the Galois group of $x^{4}+7$ over $\mathbb{Q}$ and all of the intermediate extensions between $\mathbb{Q}$ and the splitting field of this polynomial. I have found that the splitting field is $\mathbb{Q}(i, \sqrt[4]{7}\sqrt{2})=\mathbb{Q}(i,\sqrt[4]{28})$.
Edit: (Justifying the splitting field stated above)
The roots of the polynomial are $\sqrt[4]{7}\big(\pm \frac{\sqrt{2}}{2}\pm i\frac{\sqrt{2}}{2}\big)$ so the spliting field is $E=\mathbb{Q}\bigg(\sqrt[4]{7}\big(\pm \frac{\sqrt{2}}{2}\pm i\frac{\sqrt{2}}{2}\big) \bigg)=\mathbb{Q}\bigg(\sqrt[4]{7}(\pm\sqrt{2}\pm i\sqrt{2} ) \bigg)$.
We have $\sqrt[4]{7}(\sqrt{2}+i\sqrt{2}) \in E$ and $(\sqrt[4]{7}(\sqrt{2}-i\sqrt{2}) \in E$, hence their sum $=2\sqrt[4]{7}\sqrt{2} \in E \implies \sqrt[4]{7}\sqrt{2} \in E$. Then we also have $\sqrt[4]{7}\sqrt{2}+i\sqrt[4]{7}\sqrt{2}-\sqrt[4]{7}\sqrt{2} = i\sqrt[4]{7}\sqrt{2} \in E \implies \frac{i\sqrt[4]{7}\sqrt{2}}{\sqrt[4]{7}\sqrt{2}}=i \in E$. So $\mathbb{Q}(i,\sqrt{4}\sqrt{7}) \subseteq E$. But all of the roots of $x^{4}+7$ are contained in $\mathbb{Q}(i,\sqrt[4]{7}\sqrt{2})$ so we in fact have $E=\mathbb{Q}(i,\sqrt[4]{7}\sqrt{2})=\mathbb{Q}(i,\sqrt[4]{28})$. 
I'm pretty confident this is correct and I have determined the Galois group is the automorphisms of the form 
$ \sigma:  \bigg\{ \begin{array}{ll}
         i \mapsto \pm i\\
        \sqrt[4]{28} \mapsto \pm i^{k}\sqrt[4]{28}& k \in \{0,1\} \end{array} $ 
which will be isomorphic to $D_{8}$, so there should be 10 intermediate subfields.
Edit 2: I think I found a systematic way to find the ten intermediate subfields.

Still trying to figure out what the last three intermediate fields are in terms of my generators. 

$\mathbb{Q}(i,\sqrt[4]{28})\\
\mathbb{Q}(i,\sqrt{28})\\
\mathbb{Q}(i)\\
\mathbb{Q}(\sqrt[4]{28})\\
\mathbb{Q}(i\sqrt[4]{28})\\
\mathbb{Q}(i\sqrt{28})\\
\mathbb{Q}\\
$
 A: As you argue, the splitting field for $x^4+7$ is $\mathbb{Q}\bigl(i,\sqrt[4]{28}\bigr)$.  Since $\bigl|\mathbb{Q}\bigl(i,\sqrt[4]{28}\bigr) :\mathbb{Q}(i)\bigr| = 4$ and $|\mathbb{Q}(i) :\mathbb{Q}| = 2$, the extension has degree $8$, so the Galois group has order $8$.  It's not hard to see that the Galois group is dihedral, and is generated by the automorphisms $r$ and $s$ defined by
$$
r(i) = i,\qquad r\bigl(\sqrt[4]{28}\bigr) = i\sqrt[4]{28},\qquad s(i)=-i,\qquad s\bigl(\sqrt[4]{28}\bigr)= \sqrt[4]{28}.
$$
Note that $s$ is just complex conjugation.
The dihedral group of order $8$ has ten different subgroups.  Here is a list of the corresponding intermediate fields:


*

*The whole group corresponds to $\mathbb{Q}$.

*The cyclic subgroup $\{1,r,r^2,r^3\}$ corresponds to $\mathbb{Q}(i)$.

*The subgroup $\{1,r^2,s,r^2s\}$ contains $s$, so the corresponding field must be real.  It is $\mathbb{Q}\bigl(\sqrt{7}\bigr)$.

*The third subgroup $\{1,r^2,rs,r^3s\}$ of order 4 must then correspond to $\mathbb{Q}\bigl(\sqrt{-7}\bigr)$.

*The subgroup $\{1,s\}$ corresponds to the real part of the splitting field, i.e. $\mathbb{Q}\bigl(\sqrt[4]{28}\bigr)$.

*The subgroup $\{1,r^2\}$ corresponds to $\mathbb{Q}\bigl(i,\sqrt{7}\bigr)$.

*The subgroup $\{1,rs\}$ corresponds to $\mathbb{Q}\bigl(\sqrt[4]{-7}\bigr)$, where $\sqrt[4]{-7}$ denotes the principle fourth root, i.e. $e^{i\pi/4}\sqrt[4]{7}$.

*The subgroup $\{1,r^2s\}$ corresponds to $\mathbb{Q}\bigl(i\sqrt[4]{28}\bigr)$.

*The subgroup $\{1,r^3s\}$ corresponds to $\mathbb{Q}\bigl(i\sqrt[4]{-7}\bigr)$.

*The trivial subgroup corresponds to the whole splitting field $\mathbb{Q}\bigl(i,\sqrt[4]{28}\bigr)$.
In each case, you can check that the listed field is correct by checking that it has the right degree and that its elements are fixed by the action of the given elements of the Galois group.

Edit: I'd like to answer the question of how one finds all these subfields.  It shouldn't take more than a few minutes to find all of them.
Given a subgroup $H$ of the Galois group $G$, one of the best ways of finding generators for the fixed field of $H$ is symmetrization.  The way that this works is that you take any element $\alpha$ of the splitting field and compute
$$
\sum_{h\in H} h(\alpha)\qquad\text{or}\qquad \prod_{h\in H} h(\alpha).
$$
Either of these quantities is invariant under the action of $H$, and therefore lies in the fixed field.
For example, to compute the fixed field for the subgroup $\{1,rs\}$ above, we start with the element $\alpha = \sqrt[4]{28}$ and symmetrize:
$$
\alpha + rs(\alpha) \;=\; \sqrt[4]{28} + i\sqrt[4]{28} \;=\; 2\sqrt[4]{-7}.
$$
Thus $\sqrt[4]{-7}$ lies in the fixed field.  Since $\mathbb{Q}\bigl(\sqrt[4]{-7}\bigr)$ has the right degree, it must be the fixed field for $\{1,rs\}$.
Similarly, to compute the fixed field for $\{1,r^2\}$, we start with $\alpha \sqrt[4]{28}$ and symmetrize multiplicatively:
$$
\alpha\,r^2(\alpha) \;=\; \sqrt[4]{28} \bigl(-\sqrt[4]{28}\bigr) \;=\; -2\sqrt{7}.
$$
Thus $\sqrt{7}$ lies in the fixed field for $\{1,r^2\}$.  Clearly $i$ also lies in this fixed field, and $\mathbb{Q}\bigl(i,\sqrt{7}\bigr)$ has the right degree, so the fixed field for $\{1,r^2\}$ is $\mathbb{Q}\bigl(i,\sqrt{7}\bigr)$.
