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: As usual with bounty questions, this has been "answered" although not yet completely solved --- further comments still welcome. 



Length of I/I^2 versus Ann(I)/Ann(I)^2 in Artinian rings.
UPDATE 10/04/10:


Here are some partial results in the graded case. Let $R=\oplus\_0^s R\_i$ be a graded Gorenstein algebra over $k=R\_0$ ($s$ is the socle degree). $R$ is said to have *strong Lefschetz property* (SLP) if for a general linear form $l$ in $R$, the multiplication map $\times l^a: R\_i \to R\_{i+a}$ has maximal rank for all $i\geq 0, a \geq 1$ (so it is either injective or surjective). Such $l$ is called a Lefschetz element.


By a result of Stanley, monomial complete intersections of characteristic $0$, e.g. $R=k[x\_1\cdots,x\_d]/(x\_1^{a\_1},\cdots, x\_d^{a\_d})$ have SLP (apply the Hard Lefschetz Theorem!). Many more classes of rings with SLP are known, and I think it is conjectured to hold for all complete intersections, at least in char. $0$ (the keywords are Weak Lefschetz and Strong Lefschetz Property, the literature is quite big, see for example [this paper](http://atlas.mat.ub.es/personals/miro/MMR3.pdf)). 


Now, suppose $R$ is Gorenstein with SLP, and $x$ be a Lefschetz element. I claim that such an element would satisfy your inequality. Let $L$ be the annihilator of $x^2$. By Fact 1 we have that $\text{length} (x/x^2) = \text{length}(L) - \text{length}(J)$. 


Let $h\_i = \text{dim}\_k R\_i$. Then since $R$ is Gorenstein with SLP, the sequence $h\_0,\cdots,h\_s$ is unimodal and symmetric. Now since the maps $\times x$ and $\times x^2$ are always injective or surjective, it is not hard to compute that $\text{length}(L)=h\_n+h\_{n+1}$ and $\text{length}(J)=h\_n$ with $n=\lfloor s/2\rfloor$. So $\text{length} (x/x^2) =h\_{n+1}$.


If $s$ is odd, then $h\_n=h\_{n+1}$ and $J$ must live in degrees at least $n+1$, thus $J^2=0$ and equality actually holds.


If $s$ is even, then $\text{length} (x/x^2)=h\_{n+1}$. Now if $h\_n= h\_{n+1}$, then as above $J^2=0$ and equality holds. If $h\_n > h\_{n+1}$, then $J$ lives only in degree $2n=s$, but $h\_s=1$ (remember $R$ is Gorenstein) so $\text{length} (J^2)$ is at most $1$. So $\text{length}(J/J^2) \geq h\_n-1\geq h\_{n+1}$. 


I think one can push this argument to show the inequality for $x=l^a$ for any Lefschetz element $l$ and $a>0$. Because the SLP is an open condition, this would imply the inequality for general forms of degree $a$. 


I also have some examples of equality in positive characteristic with pretty interesting patterns, but I need more time to think about them. 


End of UPDATE


(Some history: Yesterday when I saw this question I posted a simple solution, which I immediately realized is wrong. After a few email exchanges with FC we were both convinced that my first attempt would not work.)


I have thought about the question a bit today but could not quite prove it. Since I may not have time to work on it more in the next few days, I will put some of my thoughts here in case they help anyone. 


Let $I=(x)$ and $J$ be the annihilator of $(x)$. Also, I will use $R$ instead of $A$.



> 
> Fact 1: $(x) \cong R/J\cong D(R/J)$
> 
> 
> 


Proof: For the first isomorphism, just look at the map $R\to R$ by multiplying with $x$. Now
$D(R/J) = \text{Hom}(R/J,R)$ is isomorphic to the annihilator of $J$, which is $(x)$ again. 



> 
> Fact 2: $x/x^2 \cong R/(J+x)$.
> 
> 
> 


Proof: $x/x^2 = (x)\otimes R/(x) =R/J\otimes R/(x)$



> 
> Fact 3: $R/J$ is itself Gorenstein.
> 
> 
> 


Proof: The canonical module of $R/J$ is $D(R/J)$ which is isomorphic to $R/J$ by Fact 1.



> 
> Fact 4: Let $(S,m,k)$ be a Gorenstein, artinian local ring. An $S$-module map $S\to M$ is injective iff the image of the generator of the socle of $S$ (which is $\text{Hom}(k,S)$ and is 1-dimensional as $S$ is Gorenstein) is non-zero. 
> 
> 
> 


Proof: If the kernel is some non-zero ideal $K$, then some element in $K$ would have $m$ as the annihilator. But then such element has to be inside the socle of $S$, which is a 1-dim vector space. 



> 
> Fact 5: For any ideal $L$, $L/L^2 = \text{Tor}\_1(R/L,R/L)$
> 
> 
> 


Proof: tensor $0\to L \to R\to R/L$ with $R/L$.


Here are a couple of approaches I tried with some comments:


**Random thoughts A:**
By Fact 2 we need to prove $$\text{length}(J) -\text{length}(J^2)\geq \text{length}(R)- \text{length}(J+x) $$
Rearranging, one needs to prove $$\text{length}(R/J) \leq \text{length}((J+x)/J^2)$$ 
Let $S=R/J$ and $M=(J+x)/J^2$. One obvious thing to try is to show that $S$ (which is Gorenstein by Fact 3) can embed in $M$. Let $s\in R$ be a lift of the socle generator of $S$. By Fact 4, for a counter example one would need 
$$(J+x)s \subseteq J^2 $$ 


This rules out many potential counter examples because of degree reasons, or if $J^2$ is too small. Note that $xs$ represents the socle generator of $R$.


**Random thoughts B:** Here is a formulation that only involves $x$. By Fact 5 we need to prove $$\text{length}(\text{Tor}\_1(R/(x),R/(x)) \leq \text{length}(\text{Tor}\_1(R/J,R/J)$$
Which, by Fact 1 is really: 
$$\text{length}(\text{Tor}\_1(R/(x),R/(x)) \leq \text{length}(\text{Tor}\_3(R/(x),R/(x))$$


This equivalent statement actually makes me a little doubtful. It might be true, and sort of make sense, because the free resolution of $R/(x)$ will typically gets bigger and bigger, but many similar homological statements about Gorenstein rings turn out to be false (although it is often not easy to cook up examples). 


So I would say that the statements is likely to be true for small (in terms of lengths or degrees) rings, because of Thoughts A, but might be false in general. Of course, I would be very happy to be wrong, and may be I was missing something really simple. 


PS: thanks for asking a nice question in commutative algebra (-: 


