Let $a = \sqrt{\sqrt2 + \sqrt3}$ Is $\mathbb{Q}(a)$ a normal extension of $\mathbb{Q}$?
I thought the answer was no because the minimum polynomial of $a$ is $x^8 -10x^4 + 1$ so $[\mathbb{Q}(a):\mathbb{Q}] = 8$ and I read a similar question here where they say that is a normal extension iff the order of the Galois group is 4, but wouldn't this mean the order of the Galois group is 8? Also, wouldn't an extension by $a$ not include anything imaginary and therefore only half of the other roots. I would almost definitely conclude that this extension is not normal, but part b of the question is to describe the Galois group $\operatorname{Gal}(\mathbb{Q}(a)/\mathbb{Q})$ up to isomorphism, which would mean that the extension would have to be normal
The normal closure of $\Bbb Q(\sqrt{\sqrt2+\sqrt3})$ contains a square root $\alpha$ of $\sqrt2-\sqrt3$. As $\alpha^2<0$, $\alpha\notin\Bbb R$. As $\Bbb Q(\sqrt{\sqrt2+\sqrt3})\subseteq\Bbb R$, then $\Bbb Q(\sqrt{\sqrt2+\sqrt3})$ is not a normal extension of $\Bbb Q$.