Problem: Let $\alpha$ be a real number such that $\alpha^4=5$. Then show that $\mathbb Q(\alpha +i\alpha) $ over $\mathbb Q$ is not a normal extension.(where $i^2=1$)
My approach: I could show that the minimal polynomial of $(\alpha+i\alpha)$ over $\mathbb Q$ is $x^4+20$ and in the splitting field it splits as $$x^4+20=(x-(\alpha+i\alpha))(x+(\alpha+i\alpha))(x-(\alpha-i\alpha))(x+(\alpha-i\alpha))$$$$=(x-(\alpha+i\alpha))(x+(\alpha+i\alpha))(x^2+2i\alpha^2)$$
As $(\alpha+i\alpha)\in \mathbb Q(\alpha+i\alpha)$ we have $-(\alpha+i\alpha)\in \mathbb Q(\alpha+i\alpha)$.
If I can show that $(x^2+2i\alpha^2)$ is irreducible over $\mathbb Q(\alpha+i\alpha)$ then I am done, that is $(\alpha-i\alpha)\not\in\mathbb Q(\alpha+i\alpha)$ which is equivalent to showing $i\not\in Q(\alpha+i\alpha)$ $\left( \because \frac{(\alpha+i\alpha)}{(\alpha-i\alpha)}=-i\right)$
But I could not figure that out. I am stuck here. I don't know if I have to use Galois' Correspondence here. Help me out. Thanks in advance!
Other approaches are welcome as well.
If $i \in \mathbb Q(\alpha + i\alpha)$, then $1 + i \in \mathbb Q(\alpha + i\alpha)$, and hence $\alpha = \frac{\alpha + i \alpha}{1 + i} \in \mathbb Q(\alpha + i\alpha)$ too.
So $\mathbb Q(\alpha + i \alpha) = \mathbb Q(\alpha, i)$.
However, $$[\mathbb Q(\alpha, i) : \mathbb Q] = [\mathbb Q(\alpha, i) : \mathbb Q(\alpha)] \times [\mathbb Q(\alpha) : \mathbb Q] = 2 \times 4 = 8,$$ while $$ [\mathbb Q(\alpha + i \alpha) : \mathbb Q] = {\rm deg}(X^4 + 20) = 4,$$ which is a contradiction.