I am preparing for an exam and this problem was in the recommended review questions from the textbook, but I am having trouble with it.
Let $u = e^{\pi i/q}$, where $q$ is an odd prime. Show that gal($(\mathbb{Q}(u): \mathbb{Q}) \simeq C_{q-1}$ (the cyclic group of order $q-1$).
Hint: show that $\mathbb{Q}(u^2) = \mathbb{Q}(u)$.
I think I can figure out how to prove the statement (based on an example we have previously seen) once I have shown the hint. However, I am not sure how to proceed with that part.
This follows from the group theory result that $Aut(C_p)=C_{p-1}$ for $p$ prime, which makes it clear that the Galois group contains a $C_{p-1}$ subgroup, combined with the fact that adjoining one of the roots already gives the splitting field (since the others are its powers), so the Galois group has order $p-1$ so it must equal $C_{p-1}$.