Question: Let $\alpha$ be a real root of $x^4-5 \in \mathbb{Q}[X].$ Show that $\mathbb{Q}(\alpha+i\alpha)/\mathbb{Q}$ is not Galois extension.
To show that an extension is not Galois, we just need to show that $\mathbb{Q}(\alpha+i\alpha)$ is not a splitting field of any polynomial over $\mathbb{Q}.$
So I guess we start with supposing it is the splitting field of some polynomial. But I do not know what contradiction to be derived here.
Any hint is appreciated.
The splitting field of $K$ of $X^4-5$ is $\Bbb Q(i,\sqrt[4]5)$. Its Galois group is dihedral order $8$, generated by $\sigma$ and $\tau$, where $\sigma(i)=i$ and $\sigma(\sqrt[4]5)=i\sqrt[4]5$, and where $\tau$ is complex conjugation.
Set $\alpha=\sqrt[4]5$. Then $\sigma(\tau((1+i)\alpha))=\sigma((1-i)\alpha) =(1+i)\alpha$. Therefore $\beta=(1+i)\alpha$ is fixed by $\sigma\tau$. The powers of $\tau$ take $\beta$ to four distinct values, so the stabiliser of $\beta$ under the action of $G$ is $H=\{\text{id},\sigma\tau\}$. Therefore $K$ is the fixed field of $H$. But $H$ is not a normal subgroup of $G$. Therefore $K$ is not Galois over $\Bbb Q$.