A Relation between discriminant of individual fields vs the discriminant of their Composite Let $L_{1}, \ldots, L_{m}$ be field extension of a number field $K$ and $L$ be the composite field of $L_{1}, \ldots, L_{m}$. Let $d$ be the relative discriminant of $L$ (over $K$) and $d_{1}, \ldots, d_{m}$ be the respective relative discriminants of $L_{1}, \ldots, L_{m}$. Then :
$$d \mid \prod_{i=1}^{m} d_{i}^{[L : L_{i}]}.$$
One place where I have seen index of the form $[L : L_{i}]$ come up is when we related discriminants over a tower of fields. For example, when $K \subseteq F \subseteq L$ is a tower of number fields then we have $$d_{L/K} = N_{F/K}(d_{L/F}) \cdot d_{F/K}^{[L : F]}.$$
It appears that if I replace $F = L_{i}$ above, then for each respective towers $K \subseteq L_{i} \subseteq L$ we obtain $$d = d_{L/K} = N_{L_{i}/K}(d_{L/L_{i}}) \cdot d_{L_{i}/K}^{[L : L_{i}]} = N_{L_{i}/K}(d_{L/L_{i}}) \cdot d_{i}^{[L : L_{i}]}.$$
From this, it seems that only way $d \mid \prod_{i=1}^{m} d_{i}^{[L : L_{i}]}$ is if none of the $d_{i}$ can be absorbed by any other $N_{L_{j}/K}(d_{L/L_{j}})$, which is some sort of coprime property? Does one need to impose some extra conditions to obtain the result? Thank you.
 A: An induction argument shows that we may assume $m=2$.
The mentioned transitivity of the discriminants is indeed useful here, although I will use the stronger fact that the differents are transitive: We have ${\cal D}_{L/K}={\cal D}_{L/L_1}{\cal D}_{L_1/K}$.
Claim: ${\cal D}_{L_2/K}\subseteq{\cal D}_{L/L_1}$.
Let $x\in {\cal O}_{L_2}$. Then we want to show $\delta_{L_2/K}(x)\in {\cal D}_{L/L_1}$. If $L_2\ne K(x)$ there is nothing to show as $\delta_{L_2/K}(x)=0$ in this case, so assume $L_2=K(x)$, so that $\delta_{L_2/K}(x)=m_{x,K}'(x)$ where $m$ denotes the minimal polynomial of the element over the given field. Note that we have $m_{x,L_1}\mid m_{x,K}$ in ${\cal O}_{L_1}[t]$ which implies $\delta_{L/L_1}(x)=m_{x,L_1}'(x)\mid m_{x,K}'(x)=\delta_{L_2/K}(x)$ in ${\cal O}_{L}$. Thus $\delta_{L_2/K}(x)\in {\cal D}_{L/L_1}$.
This implies ${\cal D}_{L/K}\mid {\cal D}_{L_1/K}{\cal D}_{L_2/K}{\cal O}_{L}$. Taking the norm $N_{L/K}$ and using its transitivity then gives us $$d_{L/K}\mid d_{L_1/K}^{[L:L_1]}d_{L_2/K}^{[L:L_2]}$$
