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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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Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many would consider your post rude because it is a command ("Show..."), not a request for help, so please consider rewriting it. – Zev Chonoles Feb 7 '13 at 13:55
Also, please take a look at what I changed to improve the formatting of your question (click the "edit" button). – Zev Chonoles Feb 7 '13 at 13:58
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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