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Let $K$ be a number field of degree $d$, and let $B = \{b_1,\ldots,b_d\}$ be a subset of $K$ of cardinality $d$ such that the matrix $(\mathrm{Tr}(b_ib_j))^d_{i,j=1}$ has non-zero determinant $\Delta_{B}(K)$.

  i) Show that $B$ is a basis for $K$ as a vector space over $\mathbb{Q}$.

  ii) Show if the elements of $B$ are integral and $\Delta_{B}(K)$ is square-free, then $\mathcal{O}_K = \mathbb{Z}b_1 + \cdots + \mathbb{Z}b_d$.

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This note is relevant. (Page28 around) – awllower Feb 8 '13 at 14:21
up vote 1 down vote accepted

In short: if ${\cal O}_K\supseteq\Gamma^\prime\supseteq\Gamma$ are lattices then $$ \Delta_{\Gamma}=\Delta_{\Gamma^\prime}[\Gamma^\prime:\Gamma]^2. $$ Thus if $\Delta_{\Gamma}$ is square-free, $\Gamma$ cannot be contained properly in any sublattice of ${\cal O}_K$ and so it must be ${\cal O}_K$.

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