Definition 3.4.1. For \(P \in \mathbb{P}_F\) let \(\mathcal{O}_P^{\prime}:=\operatorname{ic}_{F^{\prime}}\left(\mathcal{O}_P\right)\) denote the integral closure of \(\mathcal{O}_P\) in \(F^{\prime}\). Then the set
\[\mathcal{C}_P:=\left\{z \in F^{\prime} \mid \operatorname{Tr}_{F^{\prime} / F}\left(z \cdot \mathcal{O}_P^{\prime}\right) \subseteq \mathcal{O}_P\right\} \]is called the complementary module over \(\mathcal{O}_P\).
Proposition 3.4.2 With notation as in Definition 3.4.1 the following hold:
(a) \(\mathcal{C}_P\) is an \(\mathcal{O}_P^{\prime}\)-module and \(\mathcal{O}_P^{\prime} \subseteq \mathcal{C}_P\).
(b) If \(\left\{z_1, \ldots, z_n\right\}\) is an integral basis of \(\mathcal{O}_P^{\prime}\) over \(\mathcal{O}_P\), then
where \(\left\{z_1^*, \ldots, z_n^*\right\}\) is the dual basis of \(\left\{z_1, \ldots, z_n\right\}\).
(c) There is an element \(t \in F^{\prime}\) (depending on \(P\) ) such that \(\mathcal{C}_P=t \cdot \mathcal{O}_P^{\prime}\). Moreover,
and for every \(t^{\prime} \in F^{\prime}\) we have:
\[\mathcal{C}_P=t^{\prime} \cdot \mathcal{O}_P^{\prime} \Longleftrightarrow v_{P^{\prime}}\left(t^{\prime}\right)=v_{P^{\prime}}(t) \text { for all } P^{\prime} \mid P \text {. } \](d) \(\mathcal{C}_P=\mathcal{O}_P^{\prime}\) for almost all \(P \in \mathbb{P}_F\).
Proof. (a) The assertion that \(\mathcal{C}_P\) is an \(\mathcal{O}_P^{\prime}\)-module is trivial. Since the trace of an element \(y \in \mathcal{O}_P^{\prime}\) is in \(\mathcal{O}_P\), by Corollary 3.3.2, we have \(\mathcal{O}_P^{\prime} \subseteq \mathcal{C}_P\).
虽然直接验证是平凡的，但是我们还是有办法更快更有启发性的看出 \(\mathcal{C}_P\) 是 \(\mathcal{O}_P^{\prime}\)-模。
加法封闭性是因为：
\((1)\) \(\subseteq \mathcal{O}_P\)
\((2)\) \(\cdot \mathcal{O}_P^{\prime}\) 和 \(\operatorname{Tr}_{F^{\prime} / F}\) 关于加法都是同态(线性)运算。
考虑这样一个结构就能看清我想表达的意思： \(\left\{z \in F^{\prime} \mid \left(z \cdot \mathcal{O}_P^{\prime}\right) \subseteq \mathcal{O}_P\right\}\) ，这个结构对加法仍然是封闭的，所以本质原因就是它被定义在了一个加法封闭结构 \(\mathcal{O}_P\) 里，在这个条件下，不管外面套多少层加法同态(线性)运算，这个结构对加法都是封闭的。
模(数)乘运算封闭性是因为：
\((1)\) \(\cdot \mathcal{O}_P^{\prime}\) 运算摆在这个定义式子里就会使得满足要求的 \(z\) 乘 $ \mathcal{O}_P^{\prime}$ 还是在这种集合里。考虑结构 \(\left\{z \in F^{\prime} \mid \left(z \cdot \mathcal{O}_P^{\prime}\right) \subseteq \mathcal{O}_P\right\}\) ，这个意思就更明显了。所以作者直接都不验证就说这是显然的。
(b) First we consider an element \(z \in \mathcal{C}_P\). As \(\left\{z_1^*, \ldots, z_n^*\right\}\) is a basis of \(F^{\prime} / F\), there are \(x_1, \ldots, x_n \in F\) with \(z=\sum_{i=1}^n x_i z_i^*\). Since \(z \in \mathcal{C}_P\) and \(z_1, \ldots z_n \in \mathcal{O}_P^{\prime}\), it follows that \(\operatorname{Tr}_{F^{\prime} / F}\left(z z_j\right) \in \mathcal{O}_P\) for \(1 \leq j \leq n\). Now
\[\begin{aligned} \operatorname{Tr}_{F^{\prime} / F}\left(z z_j\right) & =\operatorname{Tr}_{F^{\prime} / F}\left(\sum_{i=1}^n x_i z_i^* z_j\right) \\ & =\sum_{i=1}^n x_i \cdot \operatorname{Tr}_{F^{\prime} / F}\left(z_i^* z_j\right)=x_j, \end{aligned} \]by the properties of the dual basis. Therefore \(x_j \in \mathcal{O}_P\) and \(z \in \sum_{i=1}^n \mathcal{O}_P \cdot z_i^*\).
Conversely, let \(z \in \sum_{i=1}^n \mathcal{O}_P \cdot z_i^*\) and \(u \in \mathcal{O}_P^{\prime}\), say \(z=\sum_{i=1}^n x_i z_i^*\) and \(u=\sum_{j=1}^n y_j z_j\) with \(x_i, y_j \in \mathcal{O}_P\). Then