Find all the intermediate fields of the splitting field of $x^4 - 2$ over $\mathbb{Q}$ Okay, so I mostly worked this out, and I even created lattice diagrams as shown below. But I have a specific question about finding intermediate fields, which I will ask shortly.
Let $\alpha = \sqrt[4]{2}$ and $\omega = e^{\frac{\pi}{4}i} = i$.  Then $L = \mathbb{Q}(\alpha, i)$ is the splitting field of $x^4 -2$ over $\mathbb{Q}$. Also, the Galois group $\Gamma_\mathbb{Q}(x^4 - 2) = D_8$ acts on the roots $\alpha, \alpha i, -\alpha,$ and $-\alpha i$, and is generated by rotation $\sigma$ and reflection $\tau$, where $\sigma(i) = i, \sigma(\alpha) = \alpha i$ and $\tau(\alpha) = \alpha, \tau(i) = -i$.
To find the intermediate fields between $L$ and $\mathbb{Q}$, find the subgroups of $D_8$ instead with the idea that finding subgroups is easier and better understood than finding intermediate fields. Then from the subgroups, use the Galois correspondence to get all the intermediate fields.
There are 10 subgroups of $D_8$ which must correspond to 10 intermediate fields.  Well, I pieced together 8 obvious candidates for intermediate fields, and in the end, I had to look up the other 2 which were $\mathbb{Q}(\alpha(1 + i))$ and $\mathbb{Q}(\alpha(1 - i))$. Those two seemed strange until I realized that $\sqrt{8\alpha^2 i} = \alpha(1 + i)$.
Finally, I was able to check fixed fields to verify the exact correspondence, and come up with the diagrams.
Question: Is there a systematic approach to finding and connecting up the corresponding intermediate fields once all the subgroups are known?
I'm guessing, in general and maybe in this example with $D_8$, there isn't a good, canonical way to anticipate and construct the field extensions?  The structure of groups and subgroups, as stated earlier, is easier and better understood than the structure of field extensions. Maybe this makes sense because the groups are finite and have only one operation, and fields are often infinite and have two operations.

and

UPDATE: In this lecture, Richard Borcherds clearly describes how to obtain the two nonobvious intermediate fields from the subgroups. Specifically, add the roots $\alpha$ and $\alpha i$ to fix by reflection one way, and then add the roots $\alpha$ and $-\alpha i$ to fix by reflection the other way.
 A: For $L/K$ a finite Galois extension with Galois group $G = \text{Gal}(L/K)$ we know from the Galois correspondence that intermediate fields $F$ correspond to subgroups $H \subseteq G$, with the intermediate field being $F = L^H$. So the question is whether there's a systematic way to compute the fixed subfield $L^H$.

Exercise 1: Suppose the characteristic of $K$ does not divide $|H|$. Then $L^H$ is the image of the averaging or Reynolds operator
$$L \ni x \mapsto \frac{1}{|H|} \sum_{h \in H} hx \in L^H.$$

So we can proceed by averaging every element of a basis of $L$, producing a set of elements which span $L^H$, and then finding a subset of these which is a basis. This won't always give the easiest-to-understand output but it will definitely work. In particular,
$$\alpha + i \alpha = \alpha + \sigma(\alpha)$$
and
$$\alpha - i \alpha = \alpha + (\tau \sigma \tau^{-1})(\alpha).$$

Exercise 2: For $p$ an odd prime, the cyclotomic field $\mathbb{Q}(\zeta_p)$ (where $\zeta_p = \exp \left( \frac{2\pi i}{p} \right)$) has a unique quadratic subfield. Find it, using Exercise 1 and the fact that the Galois group is $(\mathbb{Z}/p\mathbb{Z})^{\times}$, where $n \in (\mathbb{Z}/p\mathbb{Z})^{\times}$ acts by $\zeta_p \mapsto \zeta_p^n$.

If you get stuck at the very last step consult the Wikipedia article on quadratic Gauss sums. Have fun! An easier exercise you can do as a warmup is to first find the unique subfield of degree $\frac{p-1}{2}$.
