= Solution
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>.
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