<s>[INST] Normal closure of $\mathbb{Q}(\sqrt{11+3\sqrt{13}})$ over $\mathbb{Q}$ The following is a question from an undergrad course in Galois theory:

Find a normal closure $L$ of $K=\mathbb{Q}(\sqrt{11+3\sqrt{13}})$ over $\mathbb{Q}$

I know that normal extensions are splitting fields
Let: $X=\sqrt{11+3\sqrt{13}} \implies X^2=11+3\sqrt{13} \implies X^2-11=3\sqrt{13}\implies ({\frac{X^2-11}{3}})^2-13=0 $
Is this related to the splitting field?
Would the normal closure look something like: $\mathbb{Q}(\sqrt{11}, \sqrt{13})$ since $\sqrt{11}, \sqrt{13} \notin \mathbb{Q}$? I am guessing not since we have the weird embedded square root
I have not really got my head aruond these questions so would very much appreciate your guidance
 [/INST] You need to compute the zeros of $X^4-22 X^2+4=9\cdot ((\frac{X^2-11}3)^2-13)$. There is a general formula for equations of degree four which gives you
\begin{align*}
\sqrt{11+3\sqrt{13}} && -\sqrt{11+3\sqrt{13}} && \sqrt{11-3\sqrt{13}} && -\sqrt{11+3\sqrt{13}}
\end{align*}
So the normal closure of $K$ is $\Bbb Q(\sqrt{11+3\sqrt{13}}, \sqrt{11-3\sqrt{13}})$.
Since
$$
\sqrt{11+3\sqrt{13}}\cdot \sqrt{11-3\sqrt{13}}=\sqrt{11^2-9\cdot 13}=\sqrt{4}=2,
$$
you can see that 
$$
\sqrt{11-3\sqrt{13}}=\frac{2}{\sqrt{11+3\sqrt{13}}}\in K
$$
and it follows that $K$ is normal already.
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