Let $F$ be a field. There are two cases, either $char(F)=0$ or $(char(F),m)=1$ where $m$ will be specified.
$F[m]$ is defined to be splitting field of $x^m-1$.(In either case, $F[m]$ is galois extension due to separability.)
Let $F[m_1],F[m_2]$ be two such field s.t. $(m_1,m_2)=Z$.(i.e. They are coprime.)
$\textbf{Q:}$ Do I know $F[m_1]\cap F[m_2]=F$? If not, what is counter example? Note that this is not about $F=Q$. $F=Q$'s case is clear as there is Minkowski lower bound implies on discriminant which will indicate at least one prime ramify. Ramification of prime indicates discriminant non-coprime which is contradiction for $F=Q$. Other than $Q$, do I still have linearly disjointness?