# 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 $$\alpha = \sqrt{2}$$ 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.

• If you have a subgroup $H$ of $G=\text{Gal}(L/K)$ you can generate elements of $L^H$ by taking $a\in L$ and considering $$b=\sum_{\tau\in H}\tau(a).$$ If you choose $a$ wisely, you can often get a generator of $L^H$ this way. Sep 6, 2020 at 2:47
• More or less a duplicate of 1, 2, 3. Mind you, I rather think that Oscar did a better job than the other askers (not saying that the other questions would be bad), so I'm inclined to close the older ones as duplicates of this. I have the required dupehammer, but I would welcome other opinions before acting (to avoid unnecessary drama). Sep 6, 2020 at 6:52
• @Jyrki: hmm, I interpret the OP to be asking a more general question about how to find intermediate fields and not just the specific question of how to do it in this case. Sep 6, 2020 at 7:32
• @QiaochuYuan Noted (I also observed the same thing on my second reading). Unsure whether any dupeclosing is necessary. I also polled for more opinions in CURED. Sep 6, 2020 at 7:37
• Closing the others as duplicates of this. My view gained some support in CURED. Sep 6, 2020 at 15:22

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 $$\frac{p-1}{2}$$.

• Thanks, and let's see, running the basis $1, \alpha, \alpha^2, \alpha^3, i, \alpha i, \alpha^2 i, \alpha^3 i$ through the Reynolds operator for $H = \{e, \sigma\tau\}$ would give $1, \frac{1}{2}(\alpha + \alpha i), 0, \frac{1}{2}(\alpha^3 - \alpha^3 i) = -\frac{1}{2}\alpha^2 i(\alpha + \alpha i), \frac{1}{2}(\alpha^3 i - \alpha^3) = \frac{1}{2}\alpha^2 i(\alpha + \alpha i)$ which points to $\alpha +\alpha i$ as a generator since $(\alpha + \alpha i)^2 = 2\alpha^2 i$. Sep 6, 2020 at 4:32
• It shouldn't be necessary to go through an entire basis either. You could try to find a basis which is acted on by $H$, and then you'd only have to pick one element from each orbit. I seem to recall it might even always be possible to find a basis which is permuted freely by $H$... Sep 6, 2020 at 7:33
• Ah, yep, this is possible for $H = G$ so it's always possible, which is exactly the normal basis theorem: en.wikipedia.org/wiki/Normal_basis#Normal_basis_theorem Sep 6, 2020 at 7:39