I'm stuck on a detail of the following propositon:
Let $K, E$ be linearly disjoint number fields with degrees $n$ and $m$ over $\mathbb{Q}$ whose discriminants (over $\mathbb{Q}$) are relatively prime. Then $\mathfrak o_K \mathfrak o_E = \mathfrak o_{KE}$.
Everything is clear until after we establish $\mathfrak D(KE/E) = \mathfrak D(K/\mathbb{Q}) \mathfrak o_{KE}$. Taking inverses we get $D_{\mathfrak o_E}(\mathfrak o_{KE}) = D_{\mathbb Z}(\mathfrak o_K) \mathfrak o_{KE}$, where $D_{\mathbb Z}(\mathfrak o_K) = \{ x \in K : Tr_{K/\mathbb{Q}}(x \mathfrak o_K) \subseteq \mathbb{Z}\}$ is the dual module.
If $v_1, ... , v_n$ is an integral basis for $\mathfrak o_K / \mathbb{Z}$, and $v_1^{\ast}, ... , v_n^{\ast}$ is the dual with respect to the trace $Tr_{L/ \mathbb{Q}}$, of course $D_{\mathbb Z}(\mathfrak o_K) \mathfrak o_{KE} = (\mathbb{Z}v_1^{\ast} + \cdots + \mathbb{Z}v_n^{\ast}) \mathfrak o_{KE} = \mathfrak o_{KE}v_1^{\ast} + \cdots + \mathfrak o_{KE} v_n^{\ast}$. But Lang claims that this sum is actually equal to $\mathfrak o_E v_1^{\ast} + \cdots + \mathfrak o_E v_n^{\ast}$.
Why is this the case?