Module
Definition
- $M$ is called simple if it does not contain any nontrivial submodule.
Dual
Let $\mathcal{O}$ be a dedekind domain, $k$ its field of fractions and $K/k$ a finite separable field extension. Suppose $N\subset K$ is an $\mathcal{O}$-submodule.
The dual module is $$D_\mathcal{O}(N)\coloneqq {y\in K\mid \forall x\in N, \ Tr(xy)\in\mathcal{O}}$$
Properties
Suppose $N,M\subset K$ are $\mathcal{O}$- lattices.
- If $N=\langle{e_i}\rangle$ is a free $\mathcal{O}$-module of rank $n$, then $D_\mathcal{O}(N)$ is also free of rank $n$ generated by the dual basis ${f_j}$.
- $D_\mathcal{O}(N)$ is an $\mathcal{O}$-lattice.
- $D_\mathcal{O}(D_\mathcal{O}(N))=N$.
- $(D_\mathcal{O}(N):D_\mathcal{O}(M))_\mathcal{O}=(M:N)_\mathcal{O}$.
- Suppose $\mathfrak{p}\subset\mathcal{O}$ is a prime ideal. Then $(D_\mathcal{O}(N))_\mathfrak{p}=D_{\mathcal{O}_p}(N_\mathfrak{p})$.
- $D_\mathcal{O}(\mathcal{O}_K)$ is an $\mathcal{O}_k$- fractional ideal.