Suppose that $(A,\mathfrak{m})$ is a local Artinian ring. If $A$ is Gorenstein, then $A$ admits a dualizing functor on finite length modules defined by $D(M):= Hom_A(M,A)$ which preserves lengths. If $M$ is a finite length $A$ module, let $\mathrm{length}(M)$ denote the length of $M$. If $I$ is an ideal of $A$ and $J = \mathrm{Ann}(I)$, and $A$ is Gorenstein, it is not hard to show that $I = \mathrm{Ann}(J)$.
Question: Suppose that $A$ is a local Gorenstein Artinian ring, $I$ is an ideal, and $J = \mathrm{Ann}(I)$. If $I$ is principal, is it the case that $$\mathrm{length}(J/J^2) \ge^{?} \mathrm{length}(I/I^2)$$ If so, can one characterize when equality holds?
Remarks:
1. If $J$ is also principal, then (reversing $I$ and $J$) one
predicts there should be an equality. One can show that this is the case.
However, equality sometimes occurs without $J$ being principal, as can be seen from
the example:
$A = \mathbf{F}_2[x,y]/(x^3,y^3)$, $I = (x^2+y^2)$, $J = (x^2 + y^2,xy)$, with
$\mathrm{length}(I/I^2) = \mathrm{length}(J/J^2) = 4$.
2. My reasons for believing that the answer to the question is "yes" are somewhat obscure, and I would not be entirely surprised if it turns out to be false.
3. I do not know a counterexample to the claim that $\mathrm{length}(J/J^2) \ge
\mathrm{length}(I/I^2)$ even without the Gorenstein hypothesis. I would
be interested in seeing such a counterexample, if one exists (although I'm more interested in the Gorenstein case).
EDIT: NoAs usual with bounty questions, first version was correct after all..this has been "answered" although not yet completely solved --- further comments still welcome.
