Intersection of cyclotomic fields over an arbitary number field

Question

Let $K$ be an arbitrary number field and $m,n$ are two co-prime integers. Suppose we know that, $[K(\zeta_m):K]=\phi(m), [K(\zeta_n):K]=\phi(n).$ Then is it true that $[K(\zeta_{mn}):K]=\phi(mn)?.$ I can see a positive answer under certain conditions, like, if the discriminant to $K$ is co-prime to $mn,$ or if $K$ contains no proper abelian extension of $\mathbb{Q}.$ But I am not able to see if it is a general phenomenon.

Answer 1

At least the following phenomenon may ruin your day.

It may well happen that $K$ contains a non-trivial subfield of $\Bbb{Q}(\zeta_{mn})$ but does not contain non-trivial subfields of either $\Bbb{Q}(\zeta_m)$ or $\Bbb{Q}(\zeta_n)$. For example, if $m=3$, $n=5$. Then $K=\Bbb{Q}(\sqrt{-15})$ is a subfield of $\Bbb{Q}(\zeta_{15})$ because $\sqrt5\in\Bbb{Q}(\zeta_5)$ and $\sqrt{-3}\in\Bbb{Q}(\zeta_3)$. Here $$ [K(\zeta_3):K]=2=\phi(3),\quad [K(\zeta_5):K]=4=\phi(5), $$ but $$[K(\zeta_{15}):K]=4=\phi(15)/2$$ as $K(\zeta_{15})=K(\sqrt{-3},\zeta_5)$ and both $K(\sqrt{-3})/K$ as well as $K(\zeta_5)/K(\sqrt{-3})$ are quadratic.

Your discriminant condition would, of course, catch this example.