Let $f(\lambda) =\lambda^4 - 4\lambda^2 + 2 \in \mathbb{Q}[\lambda]$, let $E$ be the splitting field, find $E$ and $[E : \mathbb{Q}]$

Question

Let $f(\lambda) =\lambda^4 - 4\lambda^2 + 2 \in \mathbb{Q}[\lambda]$, and let $E$ be the splitting field of $f$. Find $E$ and $[E : \mathbb{Q}]$.

I found the hint of this question very confusing. The hint says "Show that $E$ is generated by a single root of $f(\lambda)$ in $E$". But I do as the following

Find out all roots of $\lambda^4 - 4\lambda^2 + 2$, which are $\pm \sqrt{2 \pm \sqrt{2}}$. I don't think $\sqrt{2+\sqrt{2}} \in \mathbb{Q}[\sqrt{2-\sqrt{2}}]$ and vice versa. So $E=\mathbb{Q}[\sqrt{2+\sqrt{2}}, \sqrt{2-\sqrt{2}}]$ and $[E : \mathbb{Q}] = 2$.

Seems like my solution contradicts to the hint, what is going wrong? Also how is this question related to $Gal(E/\mathbb{Q})$?

Answer 1

So you want to prove that $E=\Bbb{Q}(\sqrt{2+\sqrt2})$. A step-by-step way could be:

1. Show that $\sqrt2\in \Bbb{Q}(\sqrt{2+\sqrt2})$.
2. Check that $\sqrt{2+\sqrt2}\cdot\sqrt{2-\sqrt2}=\sqrt2$.
3. Show that $\sqrt{2-\sqrt2}\in\Bbb{Q}(\sqrt{2+\sqrt2})$.
4. Rejoice.