Is the field extension $\mathbb{Q}(\sqrt{5 + \sqrt{7}})$ over $\mathbb{Q}$ a Galois extension?

Question

Let $K= \mathbb{Q}(\sqrt{5 + \sqrt{7}}) $. Is the field extension $\mathbb{Q}(\sqrt{5 + \sqrt{7}})$ over $\mathbb{Q}$ a Galois extension?

Let $x = \sqrt{5 + \sqrt{7}} \implies x^2 = 5 + \sqrt{7} \implies (x^2-5)^2 = 7 \implies x^4 - 10x^2 +18 = 0$ Let $f(x) = x^4 - 10x^2 +18 $. $f$ is Eisenstein at 2, so $f$ is irreducible and monic. The minimal polynomial for $\sqrt{5 + \sqrt{7}}$ is then $f$.

$(x-\sqrt{5 + \sqrt{7}})(x+\sqrt{5 + \sqrt{7}})(x-\sqrt{5 - \sqrt{7}})(x+\sqrt{5 - \sqrt{7}}) = f(x)$.

So $K$ is Galois if and only if all the roots of $f$ are in $K$.

$\sqrt{5 + \sqrt{7}} \in K \implies (\sqrt{5 + \sqrt{7}})^{-1} \in K \implies \frac{\sqrt{5 - \sqrt{7}}}{\sqrt{18}} \in K \implies \frac{\sqrt{2}}{3} \sqrt{5 - \sqrt{7}} \in K$

So, $\sqrt{5 - \sqrt{7}} \in K \iff \sqrt{2} \in K$

I can't prove either. So maybe it's not Galois???

Answer 1

Hint: $K$ is Galois over $F=\Bbb{Q}[\sqrt{7}]$. Suppose $\sqrt{5-\sqrt{7}}$ belonged to $K$. What would the action of $\operatorname{Gal}(K/F)$ on $\sqrt{5+\sqrt{7}}\cdot\sqrt{5-\sqrt{7}}$ be?