Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 136 / 1 / a / i / Solution

The inverse different isIt is an ${\mathcal{O}}_L$-submodule of $L$. Choose an integral basis $e_1,\dots ,e_n$ of a finite-index free submodule of ${\mathcal{O}}_L$. Nondegeneracy of the trace pairing gives a dual $K$-basis $e_1^{\ast },\dots ,e_n^{\ast }$, and the codifferent lies between two finitely generated full ${\mathcal{O}}_K$-lattices obtained from these bases. It is therefore a fractional ${\mathcal{O}}_L$-ideal.

Every algebraic integer has integral trace, so ${\mathcal{O}}_L\subseteq {\mathfrak{D}}_{L/K}^{-1}$. Consequently its inverseis contained in ${\mathcal{O}}_L$. It is thus an integral ${\mathcal{O}}_L$-ideal, called the different ideal.