<s>[INST] Is the field extension $\mathbb{Q}(\sqrt{5 + \sqrt{7}})$ over $\mathbb{Q}$ a Galois extension? 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???  
 [/INST] 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?
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