Galois Extension and Galois Group for $Q(\sqrt{2+\sqrt{2}})$ I just need a little bit of help verifying and finishing this problem off:
Show that $\mathbb{Q}(\sqrt{2+\sqrt{2}})/\mathbb{Q}$ is a Galois Extension, and determine the structure of the Galois group.
proof:   Note that char$(\mathbb{Q})=0$, thus $\mathbb{Q}$ is perfect, that is, all irreducible polynomials in $\mathbb{Q}[x]$ are separable. Consider the polynomial $$p(x)=x^4-4x^2+2 \in \mathbb{Q}[x]$$ This is the minimal polynomial of $\sqrt{2+\sqrt{2}}$   over $\sqrt{2+\sqrt{2}}$. We also know that $$p(x)= \bigg(x-\sqrt{2+\sqrt{2}}\bigg)\bigg(x+\sqrt{2+\sqrt{2}}\bigg )\bigg(x+\sqrt{2-\sqrt{2}}\bigg)\bigg(x-\sqrt{2-\sqrt{2}}\bigg)$$ over $\mathbb{R}[x]$. So $p(x)$ is irredcible and has no repeated roots, i.e., $p(x)$ is separable. Now we show that $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is a splitting field over $\mathbb{R}$ for $p(x)\in \mathbb{Q}[x]$ in order to show that $\mathbb{Q}(\sqrt{2+\sqrt{2}})/\mathbb{Q}$ is a Galois Extension. Again for $$p(x)= \bigg(x-\sqrt{2+\sqrt{2}}\bigg)\bigg(x+\sqrt{2+\sqrt{2}}\bigg )\bigg(x+\sqrt{2-\sqrt{2}}\bigg)\bigg(x-\sqrt{2-\sqrt{2}}\bigg)$$ let $\alpha_1= \sqrt{2+\sqrt{2}}$, $\alpha_2=-\alpha_1$,  $\alpha_3=\sqrt{2-\sqrt{2}}$  and  $\alpha_4=-\alpha_3$. It is obvious that $\alpha_1,\alpha_2 \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$. To show that $\alpha_3 \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$, first consider $\alpha_3$ to be a element of $R$. Observe that $$\sqrt{2-\sqrt{2}}=\frac{\sqrt{2}\sqrt{2-\sqrt{2}}}{\sqrt{2}}=\frac{\sqrt{2}\sqrt{2-\sqrt{2}}}{\sqrt{2-\sqrt{2}}\sqrt{2+\sqrt{2}}}=\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}}$$ That is $\sqrt{2-\sqrt{2}}=\sqrt{2}(\sqrt{2+\sqrt{2}})^{-1}$. Since $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is a field, so we have that $\big((\sqrt{2+\sqrt{2}})^2-2\big)\in \mathbb{Q}(\sqrt{2+\sqrt{2}})$. However $$(\sqrt{2+\sqrt{2}})^2-2=\sqrt{2}$$ which means $\sqrt{2}\in \mathbb{Q}(\sqrt{2+\sqrt{2}})$. So $\sqrt{2-\sqrt{2}}=\sqrt{2}(\sqrt{2+\sqrt{2}})^{-1}\in \mathbb{Q}(\sqrt{2+\sqrt{2}})$ and $\alpha_i\in \mathbb{Q}(\sqrt{2+\sqrt{2}})$ for $i=1,..,4$. So $$\mathbb{Q}(\sqrt{2+\sqrt{2}})=\mathbb{Q}(\alpha_1, \alpha_2, \alpha_3, \alpha_4)$$ and so $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is a splitting field over $\mathbb{Q}$ for $p(x)$, hence $\mathbb{Q}(\sqrt{2+\sqrt{2}})/\mathbb{Q}$ is a Galois Extension. 
For the seconed part Let $G=Gal(\mathbb{Q}(\sqrt{2+\sqrt{2}})/\mathbb{Q})$. We have that $|G|=[\mathbb{Q}(\sqrt{2+\sqrt{2}}):\mathbb{Q}]=4$.
This is everything I have up to now. I am very new to Galois Theory so guidance and help will be very beneficial to me. Also if you can let me know if my proof of the Galois Extension holds.
 A: This is mainly correct. You didn't show that the polynomial is irreducible(rigorously). Use Eisenstein to show the polynomial is irreducible. So the Galois group is of order 4. There are two groups of order 4, up to isomorphism. They are $\mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$. To see which one, see if this Galois group is cyclic. 
A: Write $\alpha$ and $\beta$ for square roots of $2+\sqrt2$ and $2-\sqrt2$. Then
$\alpha^2\beta^2=2$. Let's choose $\beta$ to make $\alpha\beta=2$ (if you like real
numbers you can assume $\alpha$ and $\beta$ are positive). There is a Galois group element
with $\sigma(\alpha)=\beta$. Then $\sigma(\alpha^2-2)=\beta^2-2$, that
is $\sigma(\sqrt2)=-\sqrt2$. Then $\sigma(\alpha\beta)=\sigma(\alpha)\sigma(\beta)=\beta
\sigma(\beta)$. But also $\sigma(\alpha\beta)=\sigma(\sqrt2)=-\sqrt2=-\alpha\beta$
so that $\sigma(\beta)=-\alpha$. So $\sigma^2(\alpha)=-\alpha$ and
$\sigma^2(\beta)=-\beta$. It follows that $\sigma^2$ is not the identity, but
$\sigma^4$ is. So $\sigma$ has order $4$ in the Galois group, which also has order $4$.
The Galois group must be cyclic of order $4$.
