When is the extension $\Bbb{Q}(\sqrt{\sqrt{5} +a})/\Bbb{Q}$ Normal?

Question

For what values of $a \in \Bbb{Q}$ is the extension $\Bbb{Q}(\sqrt{\sqrt{5} +a})/\Bbb{Q}$ Normal? I know that a finite extension $L/K$ such as this is normal iff it is the splitting field of some $f \in K[x]$.

My Attempt:
I think that this is the extension generated by polynomial $x^4-2ax^2 + (a^2-5)$. If we have a field $L$ s.t. $\sqrt{\sqrt{5} +a} \in L$ then $(\sqrt{\sqrt{5} +a})^2=\sqrt{5} +a \in L$.
So $\sqrt{5} \in L$ and so $\pm \sqrt{\pm \sqrt{5} +a} \in L$.
So $\Bbb{Q}(\sqrt{\sqrt{5} +a})$ is the splitting field for polynomial $x^4-2ax^2 + (a^2-5) \in K[x]$ so $\Bbb{Q}(\sqrt{\sqrt{5} +a})/\Bbb{Q}$ is always a normal exension. Is this correct??

Answer 1

$\Bbb{Q}(\sqrt{\sqrt{5} +a})/\Bbb{Q}$ is a normal extension iff the extension is Galois iff the order of the Galois group is 4. That means the Galois group is either isomorphic to $C_2\times C_2$ or to $C_4$. Using well known criteria for Galois groups of quartic polynomials shows the first case occurs when $a^2-5$ is a square and the second case when $5(a^2-5)$ is a square.

$a=3$ is the first example of $C_4$ and $a=5$ the first example of $C_2\times C_2$. Those 2 values and $a=85$ are the only cases of $a<100$ with Galois group $C_2\times C_2$ or $C_4$.

Answer 2

The minimal polynomial can be computed with $b=\sqrt{\sqrt{5}+a}$ and $$ b^2=\sqrt{5}+a $$ so $$ b^4-2ab^2+a^2-5=0 $$ The roots are indeed $\pm\sqrt{a\pm\sqrt{5}}$. Suppose $0<a<\sqrt{5}$: then two of the roots are complex not real. Therefore the minimal polynomial doesn't split in the given extension, which is a subfield of $\mathbb{R}$ because $\sqrt{5}+a>0$.