It is well known that there is only one quadratic subfield of $\mathbb{Q}(\zeta)$, where $\zeta$ is a primitive $p$th root of unity for some prime $p$. I wonder if the same is true for $\mathbb{Q}(\alpha)$ with $\alpha$ a $n$th primitive root of unity (not necessarily prime).
Quadratic subfield of $\mathbb{Q}(\zeta)$
algebraic-number-theorygalois-extensionsgalois-theoryroots-of-unity
Best Answer
It's not true. Consider $\Bbb Q[\zeta_{12}]$. It contains both $\Bbb Q[i]$ and $\Bbb Q[\zeta_3]$, each of which have degree $2$ over $\Bbb Q$.