The <inverse different> is
$$
\mathfrak D_{L/K}^{-1}=\{x\in L:\operatorname{Tr}_{L/K}(x\mathcal O_L)\subseteq\mathcal O_K\}.
$$
It is an $\mathcal O_L$-submodule of $L$. Choose an integral basis $e_1,\ldots,e_n$ of a finite-index free submodule of $\mathcal O_L$. Nondegeneracy of the <trace form of a number field>[trace pairing] gives a dual $K$-basis $e_1^*,\ldots,e_n^*$, 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 inverse
$$
\mathfrak D_{L/K}=\{y\in L:y\mathfrak D_{L/K}^{-1}\subseteq\mathcal O_L\}
$$
is contained in $\mathcal O_L$. It is thus an integral $\mathcal O_L$-ideal, called the <different ideal>.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-136/1/a/i/solution