$\mathbb{Q}(\sqrt[4]{2}(1+i))/\mathbb{Q}$ isn't a normal extension, tower stuff...

Question

I'm trying to prove that $ \mathbb {Q} (\sqrt[4]{2}(1 + i)) /\mathbb {Q} $ is not normal. My attempt is:

Let $\alpha = \sqrt[4]{2}(1 + i) $. The polynomial \begin{equation} p (x) = x ^ 4 + 8 = (x- \alpha) (x + \alpha) (x- \sqrt[4]{- 8}) (x + \sqrt[4]{- 8}) \end{equation} has $ \alpha $ as root and is irreducible over $ \mathbb{Q} $ (Eisenstein for the translation $ p (x + 2) $). Suppose $ \mathbb{Q} (\alpha) $ is normal. So $ \pm \sqrt[4]{- 8} \in \mathbb{Q}(\alpha) $. A contradiction ...

I don't think this is right, because I factored $ p (x) $ over $ \mathbb{C} $ and not over $\mathbb{Q} (\alpha) $ ...

One more question: Is there a way to write this extension as a tower with two normal extensions??? For me, this tower would be a chain $ K \subset K_1 \subset K_2 \subset \mathbb{Q} (\alpha) $ with $ K_1 $ and $ K_2 $ normal extensions of $ \mathbb{Q} $, but I don't see how to prove or disprove it ...

Every help is welcome!

Answer 1

I'd say $\alpha$ is a root of $x^4+8$ and the normal closure of $\Bbb{Q}(\alpha)/\Bbb{Q}$ is $$\Bbb{Q}(\alpha,\overline{\alpha})=\Bbb{Q}(i,2^{1/4})$$ which has degree $8$ over $\Bbb{Q}$.

$\Bbb{Q}(\alpha)$ is a tower of two quadratic extensions.

The minimal polynomial of $2/\alpha$ is Eisenstein.