I am trying to find the degree of the splitting field of the irreducible polynomial $f(x)=x^4-14x^2+36\in\mathbb Q[x]$.
The roots of $f$ are $\left\{\pm\sqrt{7\pm\sqrt{13}}\right\}$, so the splitting field must be:
$$E=\mathbb Q\left(\sqrt{7+\sqrt{13}},\sqrt{7-\sqrt{13}}\right)$$
If I consider the following chain of extensions
$$\mathbb Q \subseteq \mathbb Q\left(\sqrt{7+\sqrt{13}}\right)\subseteq E$$
The first extension has degree $4$ because $f(x)=Irr\left(\sqrt{7+\sqrt{13}},\,\mathbb Q\right)$, so the splitting field has at least degree $4$ (and cannot be $4$ because this would mean that $$\sqrt{7-\sqrt{13}}\in\mathbb Q\left(\sqrt{7+\sqrt{13}}\right).$$
I know that the degree must divide $n!$, so it can only be $6,8,12$ or $24$. Thus, by the multiplicative formula of degrees, the extension $E\big /\mathbb Q(\sqrt{7+\sqrt{13}})$ can only have degree $2,3$ or $6$.
Is this argument correct? I got stuck here, how could I continue?
EDIT: solution
In fact, $E=\mathbb Q\left(\sqrt{7+\sqrt{13}}\right)$. To see this we show that $\sqrt{7-\sqrt{13}}$ is generated by the field $\mathbb Q\left(\sqrt{7+\sqrt{13}}\right)$.
$$\sqrt{7-\sqrt{13}} = \frac{6}{\sqrt{7+\sqrt{13}}}\in \mathbb Q\left(\sqrt{7+\sqrt{13}}\right)$$
So, the extension has degree four because is a simple extension and $f(x)=Irr\left(\sqrt{7+\sqrt{13}},\,\mathbb Q\right)$.
Best Answer
I think you will find that $\sqrt{7-\sqrt{13}}$ is in fact in the field generated by $\sqrt{7+\sqrt{13}}$. See what happens when you multiply these two radicals together.