Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R = \mathbf{C}[x_1, \ldots, x_n]$ and let $M$ be a graded $R$-module which is finite-dimensional over $\mathbf{C}$ and suppose $ 0 \leftarrow M \leftarrow R^g \leftarrow R^d \leftarrow \cdots $ is a minimal resolution of $M$. I believe that it follows that $d \geq gn$, but don't know how prove this, or derive it from standard results in commutative algebra.

The connection to Krull's PIT is that this theorem should cover the case $g = 1$.

share|cite|improve this question
I added a reference which might help with some positive answers. –  Hailong Dao Aug 2 '11 at 2:11

1 Answer 1

up vote 5 down vote accepted

Let $R=k[x,y]$, $I$ be any height $2$ ideal with at least 3 generators. Then $I$ has a resolution:

$0 \to R^a \to R^b \to I \to 0$

Counting ranks gives $b=a+1$. Dualizing the above sequence, noting that $I^* =R$ ($^*$ denotes $Hom_R(-,R)$), we get:

$0\to R \to R^b \to R^a \to Ext_R^1(I,R) \to 0$

Let $M = Ext_R^1(I,R)$. Clearly $M = Ext_R^2(R/I,R)$, so it has finite length since $R/I$ has finite length. If your conjecture is true then $b\geq 2a =2(b-1)$. But that contradicts our choice of $I$ ($b\geq 3$, for concreteness $I= (x^2,xy,y^2)$ should work).

EDIT: of course, with additional assumptions one may have some hope. The key words to search for is "Horrocks's conjecture". For example, this new paper by Dan Erman might be helpful to you. Theorem 1.2 looks particularly relevant!

share|cite|improve this answer
Thanks for the rapid reply. Concretely, if $I = (x^2,xy,y^2)$, then the constructed $M$ is the quotient of a free $R$-module on two generators $e$ and $f$ by the submodule generated by the 3 elements $xe, yf, ye-xf$. –  Dikran Karagueuzian Aug 2 '11 at 1:34
Dear Dikran, no worries. I was just about to add the concrete description! –  Hailong Dao Aug 2 '11 at 1:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.