Prove that $\mathbb{Q}(\sqrt{2+\sqrt{2}}) / \mathbb{Q}$ is a normal extension

Question

I need to prove that $\mathbb{Q}(\sqrt{2+\sqrt{2}}) / \mathbb{Q}$ is a normal extension and $\mathbb{Q}(\sqrt{2+\sqrt{2}}, i)=\mathbb{Q}(\phi)$, where $\phi^{4}=i$.

For the first part i use the theorem that says: every extension of degree 2 is normal.

Note that the polynomial
\begin{equation*}
q(x)=x^2 -(2+\sqrt{2}) \in \mathbb Q[x]
\end{equation*}
is minimal of $\gamma = \sqrt{2+\sqrt{2}}$ over $\mathbb Q$, where $\gamma$ is a root of $q(x)$. Hence
$[\mathbb{Q}(\gamma):\mathbb Q]=2$ and $\gamma$ is a root of $q(x)$ then $\mathbb{Q}(\gamma)$ is a splitting field of the polynomial $q(x)$ over $\mathbb Q$, then $\mathbb{Q}(\sqrt{2+\sqrt{2}}) / \mathbb{Q}$ is normal because every extension of degree 2 is normal.

my question is if the prove is right? and any idea to prove the second part? $\mathbb{Q}(\sqrt{2+\sqrt{2}}, i)=\mathbb{Q}(\phi)$, where $\phi^{4}=i$.

Thanks.

Answer 1

To show that $$\mathbb{Q}(\sqrt{2+\sqrt{2}}) / \mathbb{Q}$$ is a normal extension we can show that all conjugates lie in the field.

The conjugates are

• $\alpha_1 = \sqrt{2+\sqrt{2}}$
• $\alpha_2 = \sqrt{2-\sqrt{2}}$
• $\alpha_3 = -\sqrt{2+\sqrt{2}}$
• $\alpha_4 = -\sqrt{2-\sqrt{2}}$

We are done if we show that we can express $\alpha_2$ in terms of $\alpha_1$.

Note that $\alpha_1^2 - 2 = \sqrt{2}$ and $\alpha_1 \alpha_2 = \sqrt{2}$.

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For the second part, try to write $\phi$ in component form.