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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.

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Let $R=K[x,y]/(xy, x^2-y^2)$, $I=(x,y)$, then $length(J/J^2)=1$ but $length(I/I^2)=2$. –  TmobiusX Aug 23 '10 at 3:42
    
Drawing some pictures, I convinced myself that the statement is true for any monomial I in any monomial Artin ring over a field, i.e. $\mathbf{F}[x_1,\ldots,x_n] / (\prod x_i^{a_i}, \prod x_i^{b_i},\ldots)$. –  Vivek Shende Oct 3 '10 at 23:41
    
I would happily give my bounty to whoever settles this (-: –  Hailong Dao Oct 8 '10 at 20:51
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2 Answers 2

up vote 7 down vote accepted

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).

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 (-:

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Dear FC: At this point I think the inequality has a very good chance to hold if $R$ is a complete intersection. I am not so sure about $R$ just being Gorenstein. In fact, my calculation with the Lefschetz elements was initially aimed to find counter examples, since the two lengths must be close there. But I am not an expert in Artinian Gorenstein rings, so I may well be wrong here. I have some more calculations, which I will add to the answer. –  Hailong Dao Oct 7 '10 at 1:34
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The purpose of this separate answer is to address the other part of the question, when does equality happen? (also, my first answer is getting too long).

I. As FC pointed out, equality happens if $J$ is principal (this can be proved using Fact 1 in my other answer). Consider the following:

  1. $J$ is principal.
  2. $R/I$ is Gorenstein.
  3. $R/I$ is a complete intersection.

Then $(3)\Rightarrow (2) \Leftrightarrow (1)$. If $R$ is a complete intersection, then $(2)\Rightarrow (3)$.

Sketch of proof: equivalent of (1) and (2) can be proved using the Facts of my other answer. For example, since $J$ is isomorphic to the canonical module of $R/I$, $R/I$ is Gorenstein forces $J$ to be principal. The last statement follows from a cute result by Kunz that an almost complete intersection is never Gorenstein! (E. Kunz, Almost complete intersections are not Gorenstein, J. Alg. 28(1974), 111–-115)

II. By my 10/04/10 update in the answer above, if $R$ has SLP (for example when $R$ is a monomial complete intersection, and conjecturally for all complete intersections), $k$ has char. $0$ and the socle degree is odd, then equality occurs for $I$ generated by a general linear form (it may or may not happen when the socle degree is even).

For example, equality happens $I=(x+y+u+v)$ in $R=\mathbb Q[x,y,u,v]/(x^6,y^7,u^7,v^7)$ (socle degree $23$) but not for $R=\mathbb Q[x,y,u,v]/(x^7,y^7,u^7,v^7)$ (socle degree $24$). Note that for such $I$, $J$ will typically not be principal unless if the number of variables is $2$, see part (III) below.

III. Here are some examples in positive characteristic, with the help of Macaulay 2 (thanks to Branden Stone for helping me with programming). Numerical evidences suggested:

Conjecture: Let $R=\mathbb Z/(p)[x,y]/(x^N,y^N)$ and $I=(x^n+y^n)$ with $N\geq n$. Then equality happens if and only if there is an integer $k$ such that $N/n=k$ or $kp-1\leq N/n \leq kp+1$.

Remark: the first condition is not surprising. In $2$ variables, if $n$ divides $N$ then $R/I$ is a complete intersection, so by part (I), equality happens. I think this one can be proved but did not have enough motivation to go through the details. The point is that $R/J$ is Gorenstein (Fact 3 above). In $2$ variables, this is the same as complete intersection, so $R/J$ is $k[x,y]/(f,g)$ so everything can be written down explicitly. Note that the length of $R/J$ is the product of degrees of $f,g$ by Bezout theorem.

I did not see any clear pattern when you have more variables. For example, when $R=\mathbb Z/(3)[x,y,z]/(x^N,y^N,z^N)$ and $I=(x^2+y^2+z^2)$, the values of $N$ between $2$ and $100$ such that equality fails to occur is $3,9,27,33,75,81,99$. If you set the field to be $\mathbb Z/(5)$, those values will be $3,5,15,17,23,25,75,77,83,85,95,97$!

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FC: did Wiles believe this is true? –  Hailong Dao Oct 8 '10 at 20:02
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