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 $g = 1$.



Generalizing Krull's Principal Ideal Theorem to Modules
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](http://arxiv.org/abs/0902.0316) by Dan Erman might be helpful to you. Theorem 1.2 looks particularly relevant!


