Discriminant of the ring of integers of a composite field Let $K$,$L$ be two number fields and let $KL$ denote the composite field (the smallest subfield of $\mathbb{C}$ containing both $K$ and $L$). Denote respectively by $R$,$S$ and $T$ the ring of algebraic integers of $K$,$L$ and $KL$. Suppose $\left[KL:\mathbb{Q}\right]=\left[K:\mathbb{Q}\right]\left[L:\mathbb{Q}\right]$ and $T=RS$. Then,
$$\left(discT\right)=\left(discR\right)^{\left[L:\mathbb{Q}\right]}\left(discS\right)^{\left[K:\mathbb{Q}\right]}$$
This is exercise 23c from Marcus. I just don't see how this is true. Any help will be appreciated. 
 A: The given equality of degrees is equivalent to saying that $ K $ and $ L $ are linearly disjoint, that is, if we let $ F = KL $, any $ \mathbf Q $-basis of $ L $ is a $ K $-basis of $ F $. Let $ (\alpha_i) $ and $ (\beta_j) $ be integral bases for $ \mathcal O_K $ and $ \mathcal O_L $, respectively. Then, the given equality of compositums of rings of integers tells us that the $ (\alpha_i \beta_j) $ span $ T $, and by the equality of degrees we have that they are linearly independent. Therefore, they are an integral basis of $ T $. By definition, we have
$$ d_T = \det(\operatorname{Tr}_{F/\mathbf Q}(\alpha_i \beta_j \alpha_r \beta_s)) $$
Using the transitivity of the trace gives
$$ \det(\operatorname{Tr}_{F/\mathbf Q}(\alpha_i \beta_j \alpha_r \beta_s)) = \det(\operatorname{Tr}_{K/\mathbf Q}(\alpha_i \alpha_r \operatorname{Tr}_{F/K}(\beta_j \beta_s))) $$
The multiplication map by $ \beta_j \beta_s $ can be written by picking a $ K $-basis of $ F $, but by linear disjointness we may instead pick a $ L $-basis of $ \mathbf Q $. Since these elements all lie in $ L $, we see that there is an equality of traces $ \operatorname{Tr}_{F/K}(\beta_j \beta_s)) = \operatorname{Tr}_{L/\mathbf Q}(\beta_j \beta_s)) $, from which we obtain
$$ d_T = \det(\operatorname{Tr}_{K/\mathbf Q}(\alpha_i \alpha_r) \operatorname{Tr}_{L/\mathbf Q}(\beta_j \beta_s)) $$
It is a computational exercise to show that this last expression implies the claim (expand the determinant).
