Which one of $\sqrt{3},\sqrt{-3},\sqrt 5,\sqrt{-5},\sqrt{15},\sqrt{-15}$ is the element of $\mathbb Q(\zeta)$ Let $\zeta = e^{2\pi i/15}$. Which one of $\sqrt{3},\sqrt{-3},\sqrt 5,\sqrt{-5},\sqrt{15},\sqrt{-15}$  is the element of $\mathbb Q(\zeta)$ Explain your answer.
I get that using $e^{2\pi i/3}$ and $e^{2πi/5}$, you can proof that $\sqrt{-3},\sqrt 5,$ and $\sqrt{-15}$ is included inside $\mathbb Q(\zeta)$, but how about the other three? Is there a way to proof that it is not inside $\mathbb Q(\zeta)$? Thank you
 A: It's not the full solution, but here's an approach:
Once you prove that $\sqrt{-3}, \sqrt{5}, \sqrt{-15}$ are there, all you need to do is prove if $i=\sqrt{-1}$ is in $\mathbb{Q}(\zeta)$ or not. Note that the Galois group of $\mathbb{Q}(\zeta)$ is $(\mathbb{Z}/15\mathbb{Z})^\times \simeq (\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/4\mathbb{Z})$. By the fundamental theorem of Galois theory, you can match the subfields of $\mathbb{Q}(\zeta)$ with the the subgroups of the Galois group. For example, $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{5})$ are of degree $2$ over $\mathbb{Q}$, hence correspond to subgroups of index $2$ in $(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/4\mathbb{Z})$. Now match up the subfields with subgroups and see if you can get $i$ there.
Here's another less messy approach. Once again, start with observing that it suffices to prove if $i=\sqrt{-1}$ is in $\mathbb{Q}(\zeta)$ or not. Prove that the only roots of unity in $\mathbb{Q}(\zeta)$ are $\pm \zeta^j$. This will show that $i$ is not there.
