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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 $g = 1$.

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I added a reference which might help with some positive answers. – Hailong Dao Aug 2 '11 at 2:11
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!

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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

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