I am working through the exercises in Marcus's Number Fields textbook and am not sure how to proceed on Chapter 3, Exercise 34(b).
Given $K, L$ number fields, $R$ the ring of integers of $K$, and $A$ an additive subgroup of $L$, the exercises define $A^* = \{ \alpha \in L\ |\ \text{Tr}^{L}_{K}(\alpha A) \subset R \}$.
In part (a) I've proved that for any basis $\{\alpha_1, \ldots, \alpha_n\}$ of $L$ over $K$ there is a dual basis of $L$ over $K$ $\{\beta_1, \ldots, \beta_n\}$ such that $\text{Tr}^{L}_{K}(\alpha_i \beta_j)$ is 1 if $i = j$, 0 otherwise.
Part (b) asks us to show that if $A = R\alpha_1 \oplus \cdots \oplus R \alpha_n$ that $A^{*} = B$ where $B$ is the $R$-module generated by the $\beta_i$.
The suggested approach is to take $\gamma \in A^{*}$, show there exists $\beta$ such that $\text{Tr}^{K}_{L}((\gamma – \beta)A) = 0$, and that this implies $\gamma = \beta$.
Here's my approach so far:
- Let $\gamma \in A^{*}$ and define $m_i = \text{Tr}^{K}_{L}(\gamma\alpha_i)$; by assumption $m_i \in R$. Take $\beta = \sum_{i=1}^{n} m_i \beta_i$
- For $\alpha \in A$, $\alpha = r_1 \alpha_1 \oplus r_n \alpha_n$ and $\text{Tr}^{L}_{K}(\gamma\alpha) = \sum_{i = 0}^{n} r_i m_i = \text{Tr}^{L}_{K}(\beta\alpha)$.
- Therefore $\text{Tr}^{K}_{L}((\gamma – \beta)A) = 0$.
Where I'm stuck is the last step – showing this implies $\gamma = \beta$.
It's possible for an non-zero additive set to have a zero trace (in $\mathbb{Q}[\sqrt{n}]$ the set $\{\sqrt{n}, 2\sqrt{n}, \ldots\}$ has zero trace), so I'm guessing it must follow from the fact that the $\beta_i$s are a basis for $L$?
Thanks for any assistance you can give.
Best Answer
Since $A$ is a free $R$-module generated by the $\alpha_i$, each $\alpha_i \in A$. If $(\gamma - \beta)^{-1} \neq 0 \in L$, it can be written as a sum of the $\alpha_i$ with coefficients in $K$. As $K$ is the field of fractions of $R$ there is some $r \neq 0$ such that $r$ clears the denominators of the coefficients of the $\alpha_i$ and so $r(\gamma - \beta)^{-1} \in A$. Then $\text{Tr}^{L}_{K}(r(\gamma - \beta)(\gamma - \beta)^{-1}) = \text{Tr}^{L}_{K}(r) = rn$ where $n = [L : K]$. However $\text{Tr}^{L}_{K}((\gamma - \beta)\alpha) = 0$ for all $\alpha \in A$. Therefore $\gamma - \beta = 0$.