I am trying to prove rigorously that $\sqrt{6}\notin \mathbb{Q}(\sqrt{-3+\sqrt{3}})$ or that $\sqrt{6}\in \mathbb{Q}(\sqrt{-3+\sqrt{3}})$. Intuitively, I would say that the first option is the correct one, but I do not know how to justify this in a rigorous way.
Any ideas as to how to approach this problem?
Thanks in advance.
With $\alpha=\sqrt{-3+\sqrt{3}}$ we see that $\alpha$ is a root of $(X^2+3)^2-3$. This polynomial is irreducible by Eisenstein. Hence $\mathbb{Q}(\alpha)$ is of degree $4$ over $\mathbb{Q}$.
Suppose now that $\sqrt{6}\in\mathbb{Q}(\alpha)$. Since $\sqrt{3}\in\mathbb{Q}(\alpha)$ (look at $\alpha^2$) we would have that $\sqrt{2}\in\mathbb{Q}(\alpha)$. That is we would have that $\mathbb{Q}(\sqrt{2},\sqrt{3})\subseteq \mathbb{Q}(\alpha)$. But it is easy to see that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is already of degree $4$ over the rationals, and so we would have $\mathbb{Q}(\sqrt{2},\sqrt{3})= \mathbb{Q}(\alpha)$.
So suppose that we have rationals $a,b,c,d$ such that $(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})^2=-3+\sqrt{3}$. Squaring out we see that we would have that $a^2+2b^2+3c^2+6d^2=-3$, a contradiction.
Hence $\sqrt{6}\notin\mathbb{Q}(\alpha)$.
(And the Galois group of the splitting field is the dihedral group of order $8$.)