# Show that the $\mathbb{Z}$-span $\mathbb{Z}b'_1+\cdots+ \mathbb{Z}b'_d$ of $B^+$ does not depend on the choice of $B$

Let $K$ be a number field, let $\mathcal{O}_K$ be its ring of integers, and let $B = \{b_1,\ldots,b_d\}$ be a subset of $K$ of cardinality $d$ such that $\mathcal{O}_K = \mathbb{Z}b_1+\cdots+\mathbb{Z}b_d$. Define the trace-dual basis $B^+$ by the property $\mathrm{Tr}(b_ib'_j)=\delta_{i,j}$ (Kronecker delta).

i) Show that the $\mathbb{Z}$-span $\mathbb{Z}b'_1+\cdots+ \mathbb{Z}b'_d$ of $B^+$ does not depend on the choice of $B$. This abelian group is called the trace dual of $O_K$, and we denote it by $O^+_K$.

ii) Show that $\mathcal{O}^+_K$ contains $\mathcal{O}_K$, and that the discriminant $\Delta(K)$ is the index $[\mathcal{O}^+_K : O_K]$.

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Hint for i): Show that $\mathcal O_K^+=\{b'\mid \forall b\in\mathcal O_K\colon \operatorname{Tr}(bb')\in\mathbb Z\}$.