 Solution

Write $\alpha =\sqrt[n]{m}$ and $\zeta =e^{2\pi i/n}$. The $n$ field embeddings of $L$ into $\mathbb{C}$ send $\alpha$ to ${\zeta }^j\alpha$, so the field trace isFor the trace pairing $T(x,y)={\mathrm{Tr}}_{L/\mathbb{Q}}(xy)$, the basis ${\beta }_i={\alpha }^i$ therefore has dual basis

Every element of $\mathbb{Z}[\alpha ]$ is an algebraic integer. If $x\in {\mathcal{O}}_L$ and $z\in \mathbb{Z}[\alpha ]$, then $xz$ is an algebraic integer and its trace is an integer. Thus $x$ belongs to the trace-dual lattice of $\mathbb{Z}[\alpha ]$, which is generated over $\mathbb{Z}$ by the displayed dual basis. Since $1/n=m/(nm)$, every such linear combination lies in $(nm{)}^{-1}\mathbb{Z}[\alpha ]$. Hence