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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!

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2 Answers 2

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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 ?

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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.

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