Is $K=\mathbb{Q}(\sqrt{-5+\sqrt{5}}) $a Galois extension? I am wondering if the number field $K=\mathbb{Q}(\sqrt{-5+\sqrt{5}})=\mathbb{Q}(\alpha)$ is a Galois extension. I think it is and I have the following argument, but it feels like a circulair argument:
 I know that we have the subfield $\mathbb{Q}(\sqrt{5})$ so we have one automorphism $\sigma$ that sends $\sqrt{5}$ to $\sqrt{-5}$. The other roots of the minimal polynomial over $\alpha$ are:
\begin{align*}
-\alpha \\
-i\sqrt{5+\sqrt{5}} = \sigma(\alpha) \\
i \sqrt{5+\sqrt{5}} = \sigma(-\alpha)
\end{align*}
Clearly $-\alpha$ is also in $K$ and since $\sigma$ is an automorphism, $\sigma(\alpha)$ and $\sigma(-\alpha)$ must also. So we have a normal extension, hence a Galois extension. Is this a right argument or am I assuming something I cannot? 
 A: Let's start with a choice of radicals $x = i \sqrt{5 - \sqrt{5}}$. Note that
$x^2 = - (5-\sqrt{5})$ so $x^2 + 5 = \sqrt{5}$, $(x^2 + 5)^2 = 5 $, and finally
$$x^4 + 10 x^2 + 20 =0$$
an irreducible equation with roots $\pm i \cdot \sqrt{5 \pm\sqrt{5}}$. We see that the different choices of radicals gives isommorphic extensions.
We want to know whether $K =\mathbb{Q}(x)= \mathbb{Q}(i \sqrt{5 - \sqrt{5}})$ is normal, equivalently, whether $\mathbb{Q}(i \sqrt{5 - \sqrt{5}})$  contains $i \sqrt{5 + \sqrt{5}}$. Recall that $x^2 + 5 = \sqrt{5}$ so $\sqrt{5} \in K$. 
Now, $i \sqrt{5 - \sqrt{5}} \cdot  i \sqrt{5 +  \sqrt{5}}= - \sqrt{ 25 - 5} = - \sqrt{20} = - 2 \sqrt{5} \in K$, so $i \sqrt{5 +  \sqrt{5}}$ also in $K$. We conclude that $K$ is normal, hence the extension $K/\mathbb{Q}$ is Galois. 
${\bf Added:}$. The roots of the equation $x^4 + 10 x^2 + 20 =0$  are $\pm x$ and $\pm ( 2x + \frac{10}{x})$. The Galois group acts transitively on the roots. If $\sigma(x) = -x$ then $\sigma( \pm( 2x + \frac{10}{x})) = \mp ( 2x + \frac{10}{x})$, so $\sigma$ is of order $2$. More interesting, consider $\sigma$ that takes $x$ to $2x + \frac{10}{x}$. Then $\sigma(2x + \frac{10}{x}) = 2 (2x + \frac{10}{x}) + \frac{10}{2x + \frac{10}{x}} $  and 
$$ 2 (2x + \frac{10}{x}) + \frac{10}{2x + \frac{10}{x}} - (-x) = \frac{5 (10  x^4+ 10 x^2 + 20)}{x ( x^2+5)}=0$$
so $\sigma^2(x) = -x$. Hence, this $\sigma$ is of order $4$, and since the Galois group is of order $4$, it will be cyclic with this $\sigma$ as generator. 
A: The minimal polynomial of $\alpha = i\sqrt{5-\sqrt{5}}$ over $\mathbb{Q}$ is $(x^2+5)^2-5 = x^4+10x^2+20 $ and the conjugated roots are $\pm i\sqrt{5\pm\sqrt{5}}$. Since
$$ -i\sqrt{5+\sqrt{5}} = \frac{\sqrt{20}}{i\sqrt{5-\sqrt{5}}} = \frac{2\sqrt{5}}{\alpha} = 2\cdot\frac{\alpha^2+5}{\alpha} = 2\alpha+\frac{10}{\alpha}$$
the extension is clearly normal, hence a Galois extension.
