Find a normal closure $L$ of $K=\mathbb{Q}(\sqrt{10+3\sqrt{13}})$ and describe the structure of $\mathrm{Gal}(L/\mathbb{Q})$
My idea is that the normal closure is $L=\mathbb{Q}(\sqrt{10+3\sqrt{13}}, \zeta)$ where $\zeta$ is primitive root of unity.
And would we find the structure by looking at the automorphisms $\tau $ which generate $\mathrm{Gal}(L/\mathbb{Q})$?
Many thanks for your help.
To calculate the normal closure you should first know the other roots of the minimalpolynomial of $\alpha= \sqrt{10+3\sqrt{13}}$. We get $\pm\sqrt{10\pm3\sqrt{13}}$. It is clear that $-\alpha \in \mathbb{Q}(\alpha)$. So you have to check if the other two roots are alsp lying in $\mathbb{Q}(\alpha)$. If this is the case $\mathbb{Q}(\alpha)$ is normal over $\mathbb{Q}$. If not you need to also adjoin the one of the other roots. Taking some root of unity (which one actually?) does not seem to be right.
You will get that, since $\beta = \sqrt{10-3\sqrt{13}}$ is complex it is not lying in $\mathbb{Q}(\alpha)$. So the normal closure $\mathbb{Q}(\alpha, \beta)$ has index 2 over $\mathbb{Q}(\alpha)$ and therefore the Galois group has $4\cdot 2=8$ elements. It can not be cyclic, so there is no one generating automorphism. You need to define two automorphisms that generate the group. You can make some observations, for example you can check if the two generators commute, to find the structure of the Galois group.