# Galois Group of $(x^2-a)^2-b$

Let $$a,b \in\mathbb{Q},\,\ b>a^2, \,\,f=(x^2-a)^2-b \in \mathbb{Q}$$. Let $$L$$ be the splitting field of $$f$$ over $$\mathbb{Q}$$ and let $$f$$ be irreducible.

So now I want to calculate the Galois Group.

The roots of $$f$$ are $$\pm \sqrt{a \pm \sqrt{b}}$$. Let $$\omega_1 = \sqrt{a + \sqrt{b}}, \,\,\omega_2 = \sqrt{a - \sqrt{b}}$$.

You can see fast that: $$[\mathbb{Q}(\omega_1, \omega_2):\mathbb{Q}]=8$$. So this is also the order of the Galois Group.

I found the automorphism to be:

$$(\omega_1 \mapsto \omega_1, \,\omega_2 \mapsto \omega_2) = \tau_1 = \text{id}$$

$$(\omega_1 \mapsto -\omega_1, \,\omega_2 \mapsto \omega_2) = \tau_2$$

$$(\omega_1 \mapsto -\omega_1, \,\omega_2 \mapsto -\omega_2) = \tau_3$$

$$(\omega_1 \mapsto \omega_1, \,\omega_2 \mapsto -\omega_2) = \tau_4$$

$$(\omega_1 \mapsto \omega_2, \,\omega_2 \mapsto \omega_1) = \sigma_1$$

$$(\omega_1 \mapsto -\omega_2, \,\omega_2 \mapsto \omega_1) = \sigma_2$$

$$(\omega_1 \mapsto -\omega_2, \,\omega_2 \mapsto -\omega_1) = \sigma_3$$

$$(\omega_1 \mapsto \omega_2, \,\omega_2 \mapsto -\omega_1) = \sigma_4$$

Now I need to find all fields between $$K$$ and $$L$$. So first I find the subgroups.

A subgroup must have order $$1, 2, 4$$ or $$8$$.

Order 1: $$\{\text{id}\}$$

Order 2: $$\langle\{\tau_2\}\rangle,\langle\{\tau_3\}\rangle,\langle\{\tau_4\}\rangle,\langle\{\sigma_1\}\rangle,\langle\{\sigma_3\}\rangle$$

Order 4: $$\langle\{\sigma_2\}\rangle,\langle\{\sigma_4\}\rangle$$

Order 8: $$G$$ the Galois Group, so just all elements

But I don't know how to continue from now. How to find all the fields corresponding to these subgroups?

First of all you didn't get all subgroups correct. The subgroup $$\langle \sigma_2 \rangle$$ and $$\langle \sigma_4 \rangle$$ are the same because $$\sigma_2^3=\sigma_4$$, $$\sigma_4^2=\sigma_2^2$$, $$\sigma_4^3=\sigma_2$$, so they span the same subgroup.

On top of that your list of subgroups is missing two more subgroups, namely $$\langle \tau_2, \tau_3 \rangle$$ or equivalently any group that is spanned by two of the non-identity transpositions $$\tau_i$$ - this will give a subgroup with four elements that is not the same as $$\langle \sigma_2 \rangle$$. The last missing subgroup is $$\langle \sigma_1, \sigma_3 \rangle$$ that consists of the elements $$\sigma_1, \sigma_3, \tau_3$$ and $$id$$.

Now by the fundamental theorem of Galois Theory each field between $$K$$ and $$L$$ corresponds to a fixed field.

The easy ones first: The fixed field of $$\{ id\}$$ is all $$L=\mathbb Q(\omega_1, \omega_2)$$. The fixed field of Gal$$(L/K)$$ is $$K=\mathbb Q$$.

The order $$2$$ subgroups correspond to field of order $$4$$ over $$\mathbb Q$$. The subgroup $$\langle \tau_2 \rangle$$ obiously fixes $$\omega_2$$ so $$L^{\langle \tau_2 \rangle} = \mathbb Q(\omega_2)$$. Then it is also clear that $$L^{\langle \tau_4 \rangle} = \mathbb Q(\omega_1)$$. Both fields have degree $$4$$ over $$\mathbb Q$$ because $$f$$ is assumed to be irreducible so it is the minimal polynomial of both $$\omega_1$$ and $$\omega_2$$. The subgroup $$\langle \tau_3 \rangle$$ fixes $$\mathbb Q(\omega_1, \omega_2)$$ which should also have degree $$4$$ over $$\mathbb Q$$ if I am not forgetting something.

For the order $$4$$ subgroups we have to look for extensions of degree $$2$$ over $$\mathbb Q$$. For example $$\mathbb Q(\omega_1^2)$$ is fixed by the group $$\langle \tau_2, \tau_3 \rangle$$. And $$\mathbb Q(\omega_1\omega_2)$$ is fixed by $$\langle \sigma_1, \sigma_3 \rangle$$.

This way you can figure out all the fields between $$L$$ and $$K$$ - by looking which elements of $$L$$ are fixed by the homomorphisms in a given group. And you always know that if $$[L:K]=n$$ and $$\#G=k$$ you are looking for a field $$E$$ with $$[E:K] = \frac nk$$.