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So I want to show that $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is Galois over $\mathbb{Q}$ and determine its Galois group.

My thoughts are as follows:

Define $\alpha := \sqrt{2+\sqrt{2}}$. Then it is easily shown that $\alpha$ satisfies $\alpha^4-4\alpha^2+2=0$.

Define $f(x) := x^4-4x^2+2$. Then $f$ is irreducible over $\mathbb{Q}$ by Eisenstein with $p=2$.

So we have that $f$ is the irreducible polynomial for $\alpha$ over $\mathbb{Q}$.

Further $|\mathbb{Q}(\alpha):\mathbb{Q}|=4$.

For $\mathbb{Q}({\alpha})$ to be Galois, it must contain all roots of $f$.

Define $K=\mathbb{Q}(\alpha)$ for convenience.

Define $\alpha := \alpha_1$.

Since $f$ has only even powers, we know that $-\alpha := \alpha_2$ is a root, and therefore contained in $K$ since $K$ is a field.

We note that the other two roots are $\alpha_3=\sqrt{2-\sqrt{2}}$ and $\alpha_4=-\sqrt{2-\sqrt{2}}$.

So in order to show $K$ is Galois, it must be shown that $\alpha_3$ and $\alpha_4$ lie in $K$.

Now $\alpha_1^2=2+\sqrt2$ and so $\sqrt2 \in K$. Thus $-\sqrt2 \in K$ since $K$ is a field.

Can somebody explain why $\alpha_3$ and $\alpha_4$ lie in $K$?

Next we are to determine the Galois group of $K$.

Assuming $K$ is Galois, since it has degree $4$ over $\mathbb{Q}$ (shown earlier), we know that its Galois group has size $4$. There are only two groups of size $4$, namely $V_4$ and $C_4$, the Klein four group and the cyclic group of order $4$.

How do we determine which of these choice is in fact the Galois Group?

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2 Answers 2

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$\alpha^2-2=\sqrt 2\in K$ by closure of multiplication and addition.

$$\frac{\sqrt 2}{\sqrt{2+\sqrt{2}}}=\frac{\sqrt 2 \cdot\sqrt{2-\sqrt 2}}{\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt 2}}=\frac{\sqrt 2\cdot\sqrt{2-\sqrt 2}}{\sqrt{4-2}}=\sqrt{2-\sqrt 2}$$

Since $K$ is a field, it has multiplicative inverses and is closed under multiplication, so $\sqrt{2-\sqrt 2}\in K$.

We can determine the nature of $\mathrm{Gal}(K/\mathbb{Q})$ by the order of each element. If $f$ is a field automorphism of $K$ and $f(\sqrt{2+\sqrt 2})=\sqrt{2-\sqrt 2}$, then $f(\sqrt 2)=f(\alpha^2-2)=f(\alpha)^2-2=-\sqrt 2$. Therefore $$f(f(\alpha))=f\left(\sqrt{2-\sqrt 2}\right)=f\left(\frac{\sqrt{2}}{\sqrt{2+\sqrt 2}}\right)=\frac{f(\sqrt 2)}{f(\sqrt{2+\sqrt 2})}=\frac{-\sqrt{2}}{\sqrt{2-\sqrt{2}}}=-\sqrt{2+\sqrt 2}$$

Therefore $\mathrm{ord}(f)> 2$ and must divide $4=|\mathrm{Gal}(F/\mathbb{Q})|$, so $\mathrm{ord}(f)=4$. It follows that the Galois group is cyclic and abelian.

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This is not a new answer, but an addition to Tim Ratigan's answer.

If you solve the system of equations
$ax_{1}^3$  + $bx_{1}^2$ + $cx_{1}$  +  $d$ = $x_{2}$
$ax_{2}^3$  + $bx_{2}^2$ + $cx_{2}$  +  $d$ = $x_{3}$
$ax_{3}^3$  + $bx_{3}^2$ + $cx_{3}$  +  $d$ = $x_{4}$
$ax_{4}^3$  + $bx_{4}^2$ + $cx_{4}$  +  $d$ = $x_{1}$,
for
$x_{1}=+\sqrt{2 + \sqrt{2}}$
$x_{2}=+\sqrt{2 - \sqrt{2}}$
$x_{3}=-\sqrt{2 + \sqrt{2}}$
$x_{4}=-\sqrt{2 - \sqrt{2}},$
then you get
$a = 1$
$b = 0$
$c =-3$
$d = 0$.

So the relationships between the roots are:
$x_{2} = x_{1}^3 - 3x_{1} = x_{1}(x_{1}^2-3)$
$x_{3} = x_{2}(x_{2}^2-3)$
$x_{4} = x_{3}(x_{3}^2-3)$
$x_{1} = x_{4}(x_{4}^2-3).$

As you can see, the Galois group is $C_{4}$.

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