$\k$-linear functional $\lambda:\R\to \k$, called [*Frobenius form*]{}. It corresponds to $1\in\R$ in the isomorphism of . For to be an isomorphism is equivalent with $\lambda$ not vanishing on any non-zero ideal of $\R$.

Every Frobenius algebra $\R$ can be decomposed as a direct product of local Frobenius algebras $\R\cong \R_1\times\cdots \times \R_n$. Local Frobenius algebras are precisely the finite dimensional local algebras with unique minimal ideal. For a local Frobenius algebra $\R$ we denote by $\mathfrak{m}\triangleleft \R $ the maximal ideal and by $\k^\prime=\R/\mathfrak{m}$ the residue field. When $\k\cong \k^\prime$ we call the local Frobenius algebra $\R$ [ *split*]{}. We know from [@eisenbud Theorem 7.7] that a finite dimensional local algebra always contains its residue field. Thus we can consider any local Frobenius algebra $\R$ as a split local Frobenius $\k^\prime$-algebra.

Let $\R$ be a finite dimensional $\k$-algebra. The algebra $$\T^*\R:=\R^*\times \R$$ given by multiplication

$$(q_1,a_1)\cdot(q_2,a_2)=(a_2q_1+a_1q_2,a_1a_2)$$ is always a Frobenius $\k$-algebra with Frobenius form $\lambda(q,a)=q(1)$. Indeed if $(q,a)$ such that $a\neq 0$ we can find $q^\prime$ such that $q^\prime(a)\neq 0$ and so $$\lambda((q,a)\cdot (q^\prime,0))=\lambda(aq^\prime,0)=q^\prime(a)\neq 0.$$ Similarly if $q\neq 0$ there is an $a^\prime$ such that $q(a^\prime)\neq 0$ and so $\lambda((q,a)\cdot(0,a^\prime))=q(a^\prime)\neq 0$. Thus $\lambda$ is non-zero on any proper ideal in 