# Galois Group of $\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}$

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?

• – Watson Dec 26 '16 at 12:56

$\alpha^2-2=\sqrt 2\in F$ 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 $F$ is a field, it has multiplicative inverses and is closed under multiplication, so $\sqrt{2-\sqrt 2}\in F$
We can determine the nature of $Gal(F/\mathbb{Q})$ by the order of each element. If $f$ is a field automorphism on $F$ 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 $4=|Gal(F/\mathbb{Q})|$, so $\mathrm{ord}(f)=4$. It follows that the Galois group is cyclic and abelian.