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For an additive character $\psi$ of a non-Archimedean local field $F$,For the character induced from the standard character of ${\mathbb{Q}}_p$, this annihilator is the inverse different.

Write $n=[L:K]$. The field trace gives a nondegenerate $K$-bilinear trace pairingThe inverse different, or codifferent, is the trace-dual latticeIt 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 $\mathrm{Tr}(sxt)=\mathrm{Tr}(x(st))\in R$. 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 inverseis 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 $\mathrm{disc}(S/R)$ is locally generated bywhere ${\alpha }_1,\dots ,{\alpha }_n$ is a local $R$-basis of $S$. Equivalently, it is the image of the determinant of the trace pairingThis 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 ${\alpha }_1,\dots ,{\alpha }_n$ of $S$. Let ${\beta }_1,\dots ,{\beta }_n$ be its trace-dual basis, so $\mathrm{Tr}({\alpha }_i{\beta }_j)={\delta }_{ij}$; this is a basis of ${\mathfrak{D}}_{S/R}^{-1}$. If $T=(\mathrm{Tr}({\alpha }_i{\alpha }_j))$, thenThus 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 givesLocalization then proves the equality over the original Dedekind domain.

Finally, the determinant-of-pairing map identifies the invertible $R$-module $({\det }_{R}S{)}^{\otimes 2}$ with the discriminant ideal. Hence in the ideal class groupup to the harmless inverse caused by the convention used to identify invertible modules with fractional ideals. In either convention the class is a square.

The Schwartz-Bruhat space $\mathcal{S}(F)$ of a non-Archimedean local field $F$ is the vector space of locally constant, compactly supported complex-valued functions on $F$. Fix a nontrivial continuous additive character $\psi :F\to {\mathbb{C}}^{\times }$ and a Haar measure $dx$. With the sign convention required in the question, the Fourier transform over a local field is

Letbe the annihilator of the valuation ring. Translation invariance givesIndeed, the integral is the volume when the character is trivial; otherwise translation by an element on which the character is nontrivial multiplies the integral by a scalar different from one, forcing it to vanish. With the usual character of conductor ${\mathcal{O}}_F$ and the normalization $\mathrm{vol}({\mathcal{O}}_F)=1$, this becomesFor the canonical character induced from ${\mathbb{Q}}_p$, the annihilator is instead the inverse different and the displayed general formula applies.

If $g(x)=f(ax+b)$ with $a\ne 0$, the substitution $u=ax+b$ and the scaling rule $dx=|a{|}^{-1}du$ give

Every locally constant compactly supported function is a finite linear combination of characteristic functions of cosets $b+a{\mathcal{O}}_F$: compactness extracts finitely many cosets on which the function is constant. The formula just proved, together with the transform of $1_{{\mathcal{O}}_F}$, shows that the transform of each such characteristic function is again locally constant and compactly supported. Therefore the Fourier transform over a local field maps $\mathcal{S}(F)$ to itself.

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.

For an $R$-lattice $M$ in $L$, its trace-dual lattice isFor the integral closure $S$ of $R$, the trace dual $S^{\vee }$ is the inverse different.