# Galois group of a certain splitting field

Let $$f$$ be the minimal polynomial for $$\sqrt{3+\sqrt{2}}$$. Find the Galois group of the splitting field $$K$$ over $$\mathbb{Q}$$.

Here are the steps that I have taken.

1. The minimum polynomial is $$x^4-6x^2+7$$.
2. The roots of this are $$\pm \sqrt{3 \pm \sqrt{2}}$$.
3. I am guessing that the Galois group is...maybe $$\mathbb{Z}/4\mathbb{Z}$$, analogous to how $$\mathbb{Z}/4\mathbb{Z}$$ is the Galois group for $$\mathbb{Q}(\sqrt{2+\sqrt{2}})$$, but I am not sure how to show this.

Any hints appreciated, Thanks!

Let $$G=Gal(K/\mathbb{Q})$$.

Hint : $$\mathbb{Q}(\sqrt{2})$$ is contained in $$K$$, and is a Galois subextension of $$\mathbb{Q}$$. It is the fixed field of $$H=Gal(K/\mathbb{Q}(\sqrt{2}))$$; thus $$G/H \simeq Gal(\mathbb{Q}(\sqrt{2})/\mathbb{Q})\simeq \mathbb{Z/2Z}$$.

The minimal polynomial of $$\sqrt{3+\sqrt{2}}$$ over $$\mathbb{Q}(\sqrt{2})$$ is $$x^2-(3+\sqrt{2})$$, so it's pretty clear that $$H\simeq \mathbb{Z/2Z}$$. So $$G\simeq \mathbb{Z/4Z}$$ or $$(\mathbb{Z/2Z})^2$$.

Which one it is will depend on whether $$\sqrt{2}\mapsto -\sqrt{2}$$ is a square in $$G$$ or not. Can you see if it can be a square ?

Here is the technique that I finally found works. There is a theorem in Hungerford's Section V, Chapter 4, Exercise 9: That allows us to classify biquadratic quartic extensions: Should the minimal polynomial be $$x^4+ax^2+b$$, we may classify the extension as such:

1. If $$b$$ is square, then the Galois group is $$\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$$.
2. If $$b(a^2-4b)$$ is square we have $$\mathbb{Z}/4\mathbb{Z}$$.
3. If neither then we have $$D_8$$. (Dummit & Foote convention, symmetries of a square.)

The proof is not relevant to answering my question so is discarded.

An example of $$1$$ is the classic $$\mathbb{Q}(\sqrt{2},\sqrt{3})$$, which is equal to $$\mathbb{Q}(\sqrt{2}+\sqrt{3})$$, which has a minimal polynomial $$x^4-10x^2+1$$. 1, is trivially a square.

An example of $$2$$ is $$\mathbb{Q}(\sqrt{2+\sqrt{2}})$$, whose minimal polynomial is $$x^4-4x^2+2$$. Notice that $$b(a^2-4b)= 16$$.

My question is the third kind, neither $$b$$ nor $$b(a^2-4b)$$ is a square. We can proceed similar to Dummit & Foote's Exercise 16 in 14.2.

We will proceed, as the Exercise suggests by solving the polynomial and enumerating the roots, let: $$\alpha_1 = \sqrt{3+\sqrt{2}}$$, $$\alpha_2 = -\sqrt{3+\sqrt{2}}$$, $$\alpha_3 = \sqrt{3-\sqrt{2}}$$, $$\alpha_4 = -\sqrt{3-\sqrt{2}}$$. Two of these roots are real while two arent.

It is easy to check that over $$\mathbb{Q}(\sqrt{2})$$, the following automorphisms: $$\sigma:\alpha_1 \mapsto \alpha_2, \alpha_3 \mapsto \alpha_3$$ and $$\tau: \alpha_1 \mapsto \alpha_1, \alpha_3 \mapsto \alpha_4$$ define the Klein-4 group ($$V_4$$, or $$\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$$ if you prefer).

But these in turn are over $$\mathbb{Q}(\sqrt{2})$$, which is degree two, so we have a Galois group of order $$8$$ (we have to show also show that $$\mathbb{Q}(\alpha_1),\mathbb{Q}(\alpha_3)$$, and their composite is Galois, because Galois over Galois is not Galois) in our hands, which is not Abelian. The only choice are $$Q_8$$ and $$D_8$$, but only $$D_8$$ has $$V_4$$ inside of it.