Notation:
- Let $HF_{M}(d)$ and $HP_M(d)$ denote the Hilbert-function resp. the Hilbert-polynomial. $\newcommand{\reg}{\operatorname{reg}}\newcommand{\depth}{\operatorname{depth}}$
- Let $i_{\reg}(I)$ denote the index of regularity of $I$, i.e. the smallest integer s.t. $HF_{R/I}(d) = HP_{R/I}(d)$ for all $d \geq i_{\reg}(I)$.
- Let $\reg(I)$ denote the Castelnuovo-Mumford regularity of $I$.
I'm trying to prove the following statement:
If $I\subset R = k[x_1,\ldots,x_n]$ is a homogeneous ideal and $I_d = R_d$ for $d \gg 0$, then $i_{\reg}(I) = \reg(I)$.
Apparently this should follow directly from the following Corollary from Eisenbud's Book:
Corollary 4.15: Let $M$ be a finitely generated graded Cohen-Macaulay $R$-module. If $s$ is the smallest number such that $HF_M(d) = HP_{M}(d)$ for all $d \geq s$, then $s = 1 − \depth(M) + \reg(M)$.
What I did was this:
From $I_d = R_d$ follows that $HF_{R/I}(d) = 0$ for $d \gg 0$, and therefore the Hilbert-polynomial is zero. As far as I've understood, this implies that $R/I$ has krull-dimension zero.
This then implies that $\depth(R/I) = {\rm kdim}(R/I) = 0$, i.e $R/I$ is Cohen-Macaulay.
Now Corollary 4.15 implies that $i_{\reg}(I) = 1-0+\reg(R/I) \neq \reg(I)$.
What have I missed?
I think $\reg(R/I) = \reg(I) + 1$, so the result would be $2$ apart from what I'm trying to achieve.
Let $$0\to F_k\to F_{k-1}\to\cdots\to F_0\to I\to 0$$ be the minimal free resolution of $I$, with $F_i=\oplus_jR(-a_{i,j})$. Then $$0\to F_k\to F_{k-1}\to\cdots\to F_0\to S\to S/I\to 0$$ is the minimal free resolution of $S/I$.
By definition, ${\rm reg}(I)=\max_{i,j}a_{i,j}-i$. In the minimal free resolution of $S/I$, the free module $F_i$ is in the $(i+1)$-th position and the $0$-th term $S$ doesn't affect the regularity. Thus $${\rm reg}(R/I)=\max_{i,j}a_{i,j}-(i+1)=\max_{i,j}a_{i,j}-i-1={\rm reg}(I)-1.$$