# Is $\mathbb{Q}(\sqrt{2+\sqrt{-5}})$ normal over $\mathbb{Q}$

I am to show that $\mathbb{Q}(\sqrt{2+\sqrt{-5}})$ is normal over $\mathbb{Q}$ and find its Galois group.

How can it be normal? The minimum polynomial is $x^4-4x^2+9$ which has 2 distinct sets of roots: $\pm\sqrt{2+\sqrt{-5}}$ and $\pm\sqrt{2-\sqrt{-5}}$. Thus my field extension doesn't contain all the roots (its missing the latter two) and hence isn't normal. What am I missing? And how would I find the Galois group?

• Note that $\frac{1}{\sqrt {2-\sqrt{-5}}} =\frac{\sqrt {2+\sqrt{-5}}}{3}$ Oct 15, 2015 at 20:18
• You can actually denest that; Wolfram Alpha says that $\sqrt{2+\sqrt{-5}}$ equals $\frac{\sqrt{10}}2+i\frac{\sqrt2}2$. Oct 15, 2015 at 20:34

Note:$$\begin{cases} \alpha=\sqrt{2+\sqrt{-5}}\\ \beta=\sqrt{2-\sqrt{-5}} \end{cases}$$

The comment from Matho123 proves that $\alpha \beta=3$ and that the Galois group of $\mathbb Q(\alpha)$ is normal.

Regarding the Galois group $G$ of $\mathbb Q(\alpha)$, what are its elements? You have the identity $\sigma_1$.

Consider $\sigma_2$ such that $\sigma_2(\alpha)=-\alpha$. Then $$3=\sigma_2(3)=\sigma_2(\alpha \beta)=\sigma_2(\alpha) \sigma_2(\beta)=-\alpha \sigma_2(\beta)$$ Hence $\sigma_2(\beta)=-\beta$ and the order of $\sigma_2$ is equal to $2$.

By similar considerations, one can prove that the elements $\sigma_3,\sigma_4$ of the Galois group $G$ such that $\sigma_3(\alpha)=\beta$, $\sigma_4(\alpha)=-\beta$ are also of order $2$.

Finally, the Galois group has four elements and all elements (except the identity) have order $2$. So $G$ is isomorphic to the Klein four-group.

• If adjoining one root of your polynomial automatically gives you all the others, you’ve got a normal extension. Oct 16, 2015 at 1:57