$\T^* \R$. \[TR\]

Let $\R$ be a Frobenius algebra over a field $\k$ with Frobenius form $\lambda:\R\to \k$. Consider the finitely generated $\k$-algebra $\R_d:=\R[t]/(t^d)$ with $\lambda'\in \operatorname{Hom}_\k(\R_d,\k)$ defined by $$\lambda'(a_0+a_1t+\dots+a_{d-1}t^{d-1})=\lambda(a_{d-1}).$$ As every non-zero ideal $I\triangleleft \R$ intersects $t^{d-1}\R$ non-trivially, $\lambda'$ is nonzero on $I$. Thus $\R_d$ is a Frobenius $\k$-algebra.

When $d=2$ we will denote $\R_2=\R[\epsilon]/(\epsilon^2)$ by $\R[\epsilon]$. When $\R$ is a Frobenius algebra, we have an isomorphism $\R[\epsilon]\cong \T^*\R$.

Fourier transform {#Fouriergeneral}
=================

Let $\M$ be a finitely generated module over a Frobenius $\k$-algebra $\R$. We define the dual $\R$-module as

$$\M^\vee:={\rm Hom}_\R(\M,\R).$$ The $\R$-module $\M$ is then *reflexive*, namely the canonical map $\M\rightarrow(\M^\vee)^\vee$ is an isomorphism by [@hazewinkel 4.12.21(a)] as $\R$ is a *Frobenius ring*. As this is crucial for us we will prove the reflexivity property in the case where $\k$ is a finite field.

Define the *Pontryagin dual* $\R$-module of $\M$ as

$$\M^\wedge:={\rm Hom}_\Z(\M,\C^\times).$$

From now on in this section we assume that $\k$ is a finite field and $\R$ a Frobenius $\k$-algebra with $1$-form $\lambda:\R\rightarrow\k$, and that $\M$ is a finitely generated $\R$-module. We choose once and for all a non-trivial additive character $\psi:\k\rightarrow\C^\times$ and we put $\varpi:=\psi\circ\lambda:\R\rightarrow\C^\times$.

Note that the kernel of the character 