= Solution
Write $n=[L:K]$. The <field trace> gives a nondegenerate $K$-bilinear <trace pairing>
$$
(x,y)\longmapsto\operatorname{Tr}_{L/K}(xy).
$$
The <inverse different>, or codifferent, is the <trace-dual lattice>
$$
\mathfrak D_{S/R}^{-1}
=\{x\in L:\operatorname{Tr}_{L/K}(xS)\subseteq R\}.
$$
It contains $S$, because the trace of an element integral over $R$ belongs to the integrally closed ring $R$. It is stable under multiplication by $S$: if $s,t\in S$ and $x\in\mathfrak D_{S/R}^{-1}$, then $\operatorname{Tr}(sxt)=\operatorname{Tr}(x(st))\in R$. Nondegeneracy of the <trace pairing> and finite generation of $S$ show that this trace dual is a finitely generated $R$-module spanning $L$. Therefore it is a <fractional ideal> of $S$.
Its inverse
$$
\mathfrak D_{S/R}
=(\mathfrak D_{S/R}^{-1})^{-1}
=\{x\in L:x\mathfrak D_{S/R}^{-1}\subseteq S\}
$$
is the <different ideal>. Since $S\subseteq\mathfrak D_{S/R}^{-1}$, every such $x$ lies in $S$; hence $\mathfrak D_{S/R}$ is an <integral ideal> of $S$.
The <discriminant ideal> $\operatorname{disc}(S/R)$ is locally generated by
$$
\det\bigl(\operatorname{Tr}_{L/K}(\alpha_i\alpha_j)\bigr)_{1\leq i,j\leq n},
$$
where $\alpha_1,\ldots,\alpha_n$ is a local $R$-basis of $S$. Equivalently, it is the image of the determinant of the trace pairing
$$
(\det_R S)^{\otimes2}\longrightarrow R.
$$
This formulation makes the definition independent of a basis, since changing a basis multiplies its discriminant by the square of the determinant of the change-of-basis matrix.
The asserted identity of ideals can be checked after <localization> at every nonzero <prime ideal> of $R$. We may therefore assume that $R$ is a <discrete valuation ring> and choose a basis $\alpha_1,\ldots,\alpha_n$ of $S$. Let $\beta_1,\ldots,\beta_n$ be its trace-dual basis, so $\operatorname{Tr}(\alpha_i\beta_j)=\delta_{ij}$; this is a basis of $\mathfrak D_{S/R}^{-1}$. If $T=(\operatorname{Tr}(\alpha_i\alpha_j))$, then
$$
\alpha_j=\sum_i T_{ij}\beta_i.
$$
Thus the determinant measuring the inclusion $S\subseteq\mathfrak D_{S/R}^{-1}$ is $\det T$. The determinant description of the <norm of a fractional ideal> consequently gives
$$
N_{L/K}(\mathfrak D_{S/R})=(\det T)=\operatorname{disc}(S/R).
$$
Localization then proves the equality over the original <Dedekind domain>.
Finally, the determinant-of-pairing map identifies the invertible $R$-module $(\det_R S)^{\otimes2}$ with the discriminant ideal. Hence in the <ideal class group>
$$
[\operatorname{disc}(S/R)]=[\det_R S]^2,
$$
up to the harmless inverse caused by the convention used to identify invertible modules with fractional ideals. In either convention the class is a square.
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