Normal closure of $\mathbb{Q}(\sqrt{11+3\sqrt{13}})$ over $\mathbb{Q}$

Question

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

Answer 1

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.

Answer 2

Hint:

Suppose you are given a two-layer square root $\sqrt{a+\sqrt{b}}$. If $a^2-b=r^2$ for some positive rational number $r$, then:

$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$

$x=(a+r)/2$

$y=(a-r)/2$