Index of a number field and its subfields Let $F$ and $L$ be number fields, let $FL$ be their compositum, and let the discriminants of these two fields be coprime.
Given one of the extensions $F / \mathbb{Q}$ and $L / \mathbb{Q}$ is Galois I want to
show $[F L: \mathbb{Q}]=[F: \mathbb{Q}][L: \mathbb{Q}].$ I tried applying the formula for the discriminant using embeddings into $\mathbb{C}$ but got nowhere.

I don't think there is any elementary solution because it doesn't hold when replacing $\Bbb{Q}$ by another number field $K$ as there might be some unramified extension $F/K$ (so that $Disc_{F/K}(F)=O_K$) and letting $L=F$ it fails. 
A solution is that $[F L: \mathbb{Q}]=[F: \mathbb{Q}][L: \mathbb{Q}]$ is equivalent to that $F=\Bbb{Q}[x]/(f(x))$ and $f$ is irreducible over $L$. If it is not, factoring $f=gh\in L[x]$, when $F/\Bbb{Q}$ is Galois then $f$ splits completely in $F$ so that $g,h\in F[x]$ ie. $g,h\in F\cap L[x]$ and $F\cap L$ is larger than $\Bbb{Q}$. 
That the discriminants are coprime imply that the discriminant of $F\cap L$ is $(1)$ and we conclude from Minkowski theorem (that there no unramified extensions of $\Bbb{Q}$) $F\cap L=\Bbb{Q}$ ie. $f\in L[x]$ is irreducible.
