$\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is Galois over $\mathbb{Q}$? I need to show that the extension $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is Galois over $\mathbb{Q}$, and compute its Galois group. 
I am learning Galois theory by myself and got stuck in this exercise. I know the fundamental theorem of Galois theory. Any help would be useful
Thanks
 A: Let $\alpha = \sqrt{2 + \sqrt{2}}$.
An extension is Galois if it is separable and normal. Both of these properties can be resolved by considering the minimal polynomial $p$ of $\alpha$ (i.e. $p(x) = 0$).
Now $(\alpha^2 - 2)^2 - 2 = 0$ so the minimal polynomial is $x^4 - 4x^2 + 2$.
The extension is Galois since this polynomial doesn't have repeated roots. That is easy to verify by computing the gcd of $p$ with its derivative $p'$.
We can show the extension is normal by showing that $p$ splits in the field $\mathbb Q(\alpha)$. The four roots of $p$ are $\pm \sqrt{2 + \pm \sqrt{2}}$ so if we can construct $\alpha' = \sqrt{2 + \sqrt{2}}$ from $\alpha$ we would have $p(x) = (x-\alpha)(x+\alpha)(x-\alpha')(x+\alpha')$.
Observe that:


*

*$\alpha^2 - 2 = \sqrt{2}$

*$1/\alpha = \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2 + \sqrt{2}}\sqrt{2 - \sqrt{2}}} = \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2^2 - 2}} = \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2}}$


so $\alpha' = (\alpha^2 - 2)/\alpha$.
This proves that the extension is Galois.

Now to compute the Galois group. The extension can be taken in two, degree two, steps: first adjoin $\sqrt{2}$ (a root of $x^2-\sqrt{2}$) to $\mathbb Q$ then a root of $x^2-(2+\sqrt{2})$. $$[\mathbb Q : \mathbb Q(\sqrt{2+\sqrt{2}})] = [\mathbb Q : \mathbb Q(\sqrt{2})] [ \mathbb Q(\sqrt{2}) : \mathbb Q(\sqrt{2+\sqrt{2}})] = 4.$$ so by Galois theory the group too should have order 4: The group must be $C_4$ or $C_2 \times C_2$.
Next we should look at the automorphisms in detail to determine which group type this is. From Galois theory we know the group acts transitively on the roots, i.e. there is an automorphism $\sigma$ that maps $\alpha$ to any of the other roots.
Suppose $\sigma (\alpha) = \alpha'$, then $\sigma (-\alpha) = -\alpha'$ and using the relation defining $\alpha'$ compute that $\sigma (\alpha') = -\alpha$. So $\sigma$ has order for, generating $C_4$.
