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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.

For a monic degree-$n$ polynomial $g$ over a commutative ring $R$, let $A=R[X]/(g)$ and let $\lambda$ take the coefficient of $X^{n-1}$ in the reduced representative. The pairing $(a,b)\mapsto \lambda (ab)$ is unimodular on $1,X,\dots ,X^{n-1}$: reversing one basis order gives a triangular matrix with diagonal ones. When $A$ embeds in a separable field extension, Lagrange interpolation identifies $\lambda (a)$ with $\mathrm{Tr}(a/g^{\prime }(x))$. This yields the derivative formula for the inverse different.

For $R={\mathcal{O}}_k$ and $S={\mathcal{O}}_K$, define the inverse different, or codifferent, by the trace-dual latticeThe trace pairing is nondegenerate in this separable field extension, so this is a full fractional ideal of $S$. The relative different ideal is its inverse:Every integral $x$ has integral traces against $S$, so $S\subseteq {\mathfrak{D}}^{-1}$ and $\mathfrak{D}\subseteq S$.

For $k=\mathbb{Q}$, take an integral basis $b_1,\dots ,b_n$ and its trace pairing matrix $T=(\mathrm{Tr}(b_i{b}_j))$. In these coordinates the codifferent is $T^{-1}{\mathbb{Z}}^n$, andThe ideal norm of $\mathfrak{D}$ equals this index, by inversion of fractional ideal norms. Therefore the norm of the different ideal relation isHere $r_2$ is the number of conjugate pairs of complex embeddings. The different is an ideal of the upper field, whereas the relative discriminant ideal is its ideal norm in the base field; over $\mathbb{Q}$ the ordinary signed field discriminant also records the signature.

For the finite separable field extension define the inverse different, or codifferent, byIt is a fractional ideal of ${\mathcal{O}}_L$. The different ideal is its inverse fractional ideal.

Suppose the integer ring is generated by $\alpha$, with monic minimal polynomial of an algebraic element $g$ of degree $d$. We prove the derivative formula using the coefficient pairing for a monogenic algebra. If ${\alpha }_1,\dots ,{\alpha }_d$ are its distinct conjugate polynomial roots, Lagrange interpolation for a polynomial $h$ of degree less than $d$ givesComparing coefficients of $T^{d-1}$ yieldsFor any integral polynomial, first reduce modulo $g$; the coefficients stay integral because $g$ is monic over ${\mathcal{O}}_K$.

It follows that $g^{\prime }(\alpha {)}^{-1}{\mathcal{O}}_L$ lies in the codifferent. To prove equality, consider the matrixAll entries are integral. Entries with $i+j<d-1$ are zero, and those with $i+j=d-1$ are one. Reversing the column order makes the matrix triangular with diagonal ones; thus its determinant is a unit of ${\mathcal{O}}_K$.

Write an arbitrary $x\in L$ as $x=g^{\prime }(\alpha {)}^{-1}{\sum }_j{b}_j{\alpha }^j$ with $b_j\in K$. Membership in the codifferent says $A(b_j{)}_j\in {\mathcal{O}}_K^d$. Since $A$ is invertible over ${\mathcal{O}}_K$, this forces every $b_j$ integral. ThereforeSeparability ensures that $g^{\prime }(\alpha )\ne 0$; the trace pairing calculation also proves that no extra factor has been overlooked.

The inverse different is the trace-dual latticeThe integral closure ${\mathcal{O}}_L$ is finite free over the complete discrete valuation ring ${\mathcal{O}}_K$. Choose an integral basis and use the nondegenerate trace pairing to form its dual basis over $K$. This exhibits ${\mathfrak{D}}^{-1}$ as a full, finite ${\mathcal{O}}_K$-lattice. It is stable under multiplication by ${\mathcal{O}}_L$, because $b{\mathcal{O}}_L\subseteq {\mathcal{O}}_L$ for $b\in {\mathcal{O}}_L$, and bounded denominators make it a fractional ideal of ${\mathcal{O}}_L$. Integral elements have integral field traces, so ${\mathcal{O}}_L\subseteq {\mathfrak{D}}^{-1}$. A nonzero fractional ideal of a discrete valuation ring is invertible; its inverse is consequently an integral ideal, the different ideal ${\mathfrak{D}}_{L/K}$.

Assume now $A={\mathcal{O}}_K[x]={\mathcal{O}}_L$, with monic separable minimal polynomial $g$ of degree $n$. Lagrange interpolation at its distinct roots givesIndeed, these are the leading coefficients in the interpolation formula for $X^j$. For any element of $A$, this trace is the coefficient of $X^{n-1}$ in its degree-less-than-$n$ representative modulo $g$. The resulting pairing on the basis $1,x,\dots ,x^{n-1}$ is integral and unimodular: reversing the order of one basis makes its matrix triangular with diagonal ones, since entries vanish when the exponent sum is less than $n-1$. Thus it identifies $A$ with its full ${\mathcal{O}}_K$-dual. Translating back to the trace pairing proves

For a totally ramified extension of degree $n$, any uniformizer ${\pi }_L$ is an Eisenstein generator of a totally ramified extension, and ${\mathcal{O}}_L={\mathcal{O}}_K[{\pi }_L]$. To see the latter equality, use the common residue field to expand an integral element in powers of ${\pi }_L$ with digits from ${\mathcal{O}}_K$; reduce powers using its Eisenstein polynomial and take limits in the finite complete ${\mathcal{O}}_K$-module generated by $1,{\pi }_L,\dots ,{\pi }_L^{n-1}$. Write that polynomial as $X^n+{\sum }_{j<n}{a}_j{X}^j$, with $v_L(a_j)\ge n$ for $j<n$. When $p\nmid n$, the derivative's leading term has valuation $n-1$, while every other nonzero derivative term has valuation at least $n$. There can be no cancellation of the unique smallest term. Therefore

For the prime-power p-adic cyclotomic extension, put $\zeta ={\zeta }_{p^r}$ and $e=p^{r-1}(p-1)$. Modulo $p$, the shifted cyclotomic polynomial ${\Phi }_{p^r}(1+T)$ is $T^e$, while its constant term is $p$, not a multiple of $p^2$. It is therefore Eisenstein, so $\pi =\zeta -1$ is a uniformizer and ${\mathcal{O}}_{K_r}={\mathbb{Z}}_p[\zeta ]$. Fromwe obtain, by differentiating the numerator at $\zeta$,The denominator is a primitive-$p$ root of unity minus one and has $K_r$-valuation $p^{r-1}$. The different exponent is consequently $re-p^{r-1}$, givingThis includes $p=2,r=1$: the extension is trivial and its different exponent is zero.

For this non-Archimedean local field, the Schwartz-Bruhat space $\mathcal{S}(F)$ consists of locally constant, compactly supported complex-valued functions. Choose a nontrivial continuous additive character $\psi :F\to {\mathbb{C}}^{\times }$ of modulus one and an additive Haar measure $dx$. We use the plus-sign convention for the Fourier transform over a local field:Compact support makes the integral absolutely convergent. The additive character and the scale of the Haar measure are part of the definition; without them there is no canonical numerical transform.

Let $q=|{\mathcal{O}}_F/{\pi }_F{\mathcal{O}}_F|$, and write $V=\mathrm{vol}({\mathcal{O}}_F)$. Define the integer $c$ by the annihilator of the valuation ring ${\mathcal{O}}_F^{\perp }={\pi }_F^c{\mathcal{O}}_F$. Equivalently, $\psi$ is trivial on ${\pi }_F^c{\mathcal{O}}_F$ and not on ${\pi }_F^{c-1}{\mathcal{O}}_F$. Such a conductor exists: continuity puts the image of some additive ball inside an arc containing no nontrivial circle subgroup, so the character is trivial on that ball; nontriviality bounds the possible ball exponents below. Then $H_n={\pi }_F^n{\mathcal{O}}_F$ has volume $V{q}^{-n}$ and character annihilator $H_n^{\perp }={\pi }_F^{c-n}{\mathcal{O}}_F$.

Translation givesThe integral is the volume if $y\in H_n^{\perp }$. Otherwise translate it by $t_0\in H_n$ with $\psi (t_{0}y)\ne 1$; Haar measure invariance multiplies the same integral by a nonidentity scalar, so it must vanish. ThusEvery Schwartz-Bruhat function is a finite linear combination of such coset indicators: compactness of its support supplies a common sufficiently small translation subgroup on which it is constant, and finitely many of its cosets cover the support. The transform of each indicator has compact support and is locally constant, since $\psi$ has open kernel. Consequently $\hat{f}\in \mathcal{S}(F)$ for every $f\in \mathcal{S}(F)$.

Apply the transform again to an indicator. The same character-orthogonality calculation, together with $(H_n^{\perp }{)}^{\perp }=H_n$, givesChoose the self-dual Haar measure, characterized here by $V=q^{c/2}$. By linearity the desired Fourier inversion isIn particular, one may rescale any nontrivial additive character to make $c=0$, and then take $V=1$. For a concrete construction, start with ${\psi }_0(x)={\psi }_{{\mathbb{Q}}_p}({\mathrm{Tr}}_{F/{\mathbb{Q}}_p}x)$, where the standard rational additive character has kernel ${\mathbb{Z}}_p$. Its annihilator is the inverse different ${\mathfrak{D}}_{F/{\mathbb{Q}}_p}^{-1}$. If that ideal is ${\pi }_F^{-d}{\mathcal{O}}_F$, the rescaled character $\psi (x)={\psi }_0({\pi }_F^{-d}x)$ has ${\mathcal{O}}_F^{\perp }={\mathcal{O}}_F$. This supplies the stated normalization and also explains how the different ideal enters local Fourier analysis.

A normalization is important here. I use the covolume-one theta function of a fractional ideal, for which the requested limit is $1$, and also give the formula for the unnormalized Gaussian theta sum.

Let $n=[K:\mathbb{Q}]$, let $d_v=[K_v:\mathbb{R}]\in \{1,2\}$, and choose one field embedding ${\sigma }_v$ for each Archimedean place. The Minkowski embedding of a number field identifieswith a real inner product space havingIts Euclidean Lebesgue measure is $d\mu ={\prod }_{v\text{ real}}d{x}_v{\prod }_{v\text{ complex}}2d\mathrm{Re}{x}_{v}d\mathrm{Im}{x}_v$. With this convention, the covolume of a fractional ideal lattice $j(\mathfrak{b})$ iswhere $D_K$ is the field discriminant and $N(\mathfrak{b})$ is the positive absolute norm of a fractional ideal. Thus the Euclidean lattice ${\Lambda }_{\mathfrak{b}}=C_{\mathfrak{b}}^{-1/n}j(\mathfrak{b})$ has covolume $1$. DefineThis Gaussian theta sum converges absolutely whenever every $y_v>0$.

The trace dual of a fractional ideal isHere ${\mathfrak{D}}_{K/\mathbb{Q}}^{-1}$ is the inverse different. Since $N({\mathfrak{D}}_{K/\mathbb{Q}})=|D_K|$, we have $C_{{\mathfrak{b}}^{\vee }}=C_{\mathfrak{b}}^{-1}$. There is a subtle distinction between the trace pairing and the positive inner product: the dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$, where the bar conjugates the complex coordinates and fixes the real ones. Indeed,Consequently ${\Lambda }_{\mathfrak{b}}^{\ast }=\overline{{\Lambda }_{{\mathfrak{b}}^{\vee }}}$. Coordinatewise complex conjugation preserves the weighted squared lengths in the Gaussian theta sum.

Here are the precise analytic formulas used in the proof. For a Schwartz function on $K_{\infty }$, take the Fourier transform to beFor a full Euclidean lattice $\Lambda$ of covolume $C$, the Poisson summation formula for a Euclidean lattice isFor $f_y(x)=\exp (-\pi {\sum }_v{d}_v{y}_v|x_v{|}^2)$, the Gaussian Fourier transform, applied in orthonormal real coordinates, givesMore generally, $\widehat{\exp (-\pi x^{T}Ax)}(z)=(\det A{)}^{-1/2}\exp (-\pi z^T{A}^{-1}z)$ for a real positive-definite matrix $A$ that is symmetric. Each complex coordinate contributes two real coordinates, which explains its exponent $d_v=2$.

Apply the Poisson summation formula for a Euclidean lattice to ${\Lambda }_{\mathfrak{b}}$, whose covolume is $1$, and use the preceding description of its dual lattice. The anisotropic theta functional equation isFor comparison, the unnormalized Gaussian theta sumhas the functional equationand its corresponding limit is $C_{\mathfrak{b}}^{-1}$ rather than $1$. Explicitly, our normalization is $\Theta (y,\mathfrak{b})=\theta (C_{\mathfrak{b}}^{-2/n}y,\mathfrak{b})$.

For the small-parameter asymptotic of a lattice theta sum, $y\to 0$ means that every coordinate tends to zero. In $\Theta (y^{-1},{\mathfrak{b}}^{\vee })$, the zero lattice vector contributes $1$. Every nonzero vector contributes a term tending to zero. Once every $y_v\le 1$, all these terms are bounded by the summable Gaussian theta sum ${\sum }_{\lambda \in {\Lambda }_{{\mathfrak{b}}^{\vee }}}{e}^{-\pi |\lambda {|}^2}$. The dominated convergence theorem therefore gives $\Theta (y^{-1},{\mathfrak{b}}^{\vee })\to 1$. Using the anisotropic theta functional equation,The condition that every coordinate tends to zero matters; $\Vert y\Vert \to 0$ alone would not justify this argument.

For a finite extension $E/K$ of number fields, the inverse different is the fractional idealThe different ideal is its inverse. The trace pairing is nondegenerate because extensions of number fields are separable, so this definition gives a full fractional ideal. The inclusion ${\mathcal{O}}_E\subseteq {\mathfrak{D}}_{E/K}^{-1}$ shows that ${\mathfrak{D}}_{E/K}$ is integral.

Write its prime ideal factorization as ${\mathfrak{D}}_{E/K}={\prod }_{\mathfrak{P}}{\mathfrak{P}}^{d_{\mathfrak{P}}}$. For $\mathfrak{P}$ above $\mathfrak{p}$, with ramification index $e_{\mathfrak{P}}$ and residue characteristic $\ell$, the different exponent and tame ramification theorem saysThe equivalence uses the separability of finite residue-field extensions. In particular, $d_{\mathfrak{P}}=0$ precisely at unramified prime ideals; in the wild case $d_{\mathfrak{P}}\ge e_{\mathfrak{P}}$. This theorem does not require a Galois extension. The relative discriminant is the norm of the different ideal:For $K=\mathbb{Q}$, this gives $|D_E|=N({\mathfrak{D}}_{E/\mathbb{Q}})$.

Now take ${\alpha }^3=m$ and $K=\mathbb{Q}(\alpha )$. For any prime number $q\mid m$, the square-free integer hypothesis makes $X^3-m$ an Eisenstein polynomial at $q$. It is therefore irreducible, and $K$ is a pure cubic number field of degree $3$. Since $\alpha$ is an algebraic integer, $A=\mathbb{Z}[\alpha ]$ is an order in a number field. The discriminant of elements of a number field for its basis $1,\alpha ,{\alpha }^2$ isFor example, the resultant of $X^3-m$ and $3X^2$ is $27m^2$, and the degree-three sign in the polynomial discriminant is negative. If $I=[{\mathcal{O}}_K:A]$, the discriminant-index formula for an integral lattice givesOnly prime numbers dividing $3m$ can therefore divide $I$.

For $q\mid m$, the Eisenstein polynomial gives a totally ramified extension of ${\mathbb{Q}}_q$ of degree $3$. Thus $K$ has a unique prime ideal $\mathfrak{P}$ above $q$, with ramification index $3$ and residue-field degree $1$. Since $q\ne 3$, this is tame ramification, and the different exponent and tame ramification theorem gives $d_{\mathfrak{P}}=2$. Hence $v_q(D_K)=2$. Comparing with $v_q(-27m^2)=2$ in the discriminant-index formula for an integral lattice yields $v_q(I)=0$.

At $3$, use the shifted Eisenstein polynomial of $\beta =\alpha -m$:Because $3\nmid m$, the constant term is divisible by $3$. Moreover,when $3\nmid m$: the factors $m-1$ and $m+1$ cannot both be divisible by $3$. The hypotheses therefore give $v_3(m^3-m)=1$, so the translated polynomial is an Eisenstein polynomial at $3$. The resulting completion is a totally ramified extension of degree $3$, with residue-field degree $1$, and its ramification index is divisible by the residue characteristic. Thus its different exponent is at least $3$.

It follows that $v_3(D_K)\ge 3$. But $v_3(-27m^2)=3$, soforces $v_3(I)=0$ and $v_3(D_K)=3$. No prime number divides $I$. We have proved the integral basis of a nonexceptional pure cubic field:Using local Eisenstein polynomials here only establishes the local ramification indices; it does not assume that $\mathbb{Z}[\alpha ]$ is already the full ring of integers of a number field.

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.

For a nonzero fractional ideal $\mathfrak{b}$ of a number field, its trace-dual lattice is the fractional idealThe equality follows from the definition of the inverse different and invertibility of fractional ideals. Under the positive metric on the Minkowski embedding of a number field, the Euclidean dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$: coordinatewise complex conjugation converts the positive inner product into the trace pairing. This assertion does not require that conjugation preserve the number field itself.