Let $A$ be a finite-dimensional commutative algebra over a field $k$, local with maximal ideal $\mathfrak{m}$. Then it is well-known that the following are equivalent:
- $A$ has a unique minimal ideal.
- $A$ has one-dimensional socle.
- $A$ is a Frobenius algebra.
- $A$ is Gorenstein.
Then the minimal ideal is given by $\mathfrak{m}^n$ for some $n$, and is equal to the scalar multiples of a fixed element $a \in A$.
All the examples of such algebras that I'm aware of have the following duality property:
If $x \in \mathfrak{m}^j$ and $x y = 0$ for all $y \in \mathfrak{m}^{n-j}$, then $x \in \mathfrak{m}^{j+1}$.
Is this true in general under the above assumptions on $A$?
This seems to be closely related to Matils duality, but I'm not yet sure about the exact relation. I have also found literature which states and proves this under the additional assumption that $A$ is graded.
Furthermore, there's a simple induction argument reducing the problem to the following:
If $y \in A$ is such that $xy \in \mathfrak{m}^{j+1}$ for every $x \in \mathfrak{m}$, then does it follow that $y \in \mathfrak{m}^j$?
This is quite obviously false for local finite-dimensional algebras in general, but again I haven't yet found a counterexample under the above hypotheses.
Here's a counterexample. Let $R=k[x,y,z]/(xy,xz,yz,x^3-y^2,x^3-z^2)$. Then $R$ is a finite dimensional local $k$-algebra with maximal ideal $\mathfrak{m}=(x,y,z)$ and is Gorenstein. Note $\mathfrak{m}^2=(x^2)$ and the minimal ideal of $R$ is $\mathfrak{m}^3=(x^3)$. Further, $y \in \mathfrak{m}$ satisfies $yx^2=0$, but $y \notin \mathfrak{m}^2$.
The problem with this example is exactly that it is not standard graded, and furthermore the associated graded ring $\operatorname{gr}_{\mathfrak{m}}(R)$ is not Gorenstein.