David Eisenbud, Hilbert theorem

I just started reading D. Eisenbud Commutative algebra with a view towards algebraic geometry and I wonder about a theorem on page 42:

If $M$ is a finitely generated graded module over $k[x_1,...,x_r]$ then $H_M(s)$ agrees, for large $s$, with a polynomial of degree $\leq r-1$, where $H_M(s):=\dim_k(M_s)$.

I don't understand what "then $H_M(s)$ agrees, for large $s$, with a polynomial of degree $\leq r-1$" means. Can someone explain this?

It means that there is a polynomial $p(n)=a_{r-1}n^r+...+a_0$, and a number $N$ such that $H_M(s)=p(s)$ for $s>N$.

Maybe more important to understand than that statement is the following equivalent one.

Define $\Delta f(n):=f(n+1)-f(n)$ and $\Delta^{m}f(n)=\Delta\Delta^{m-1}f(n)$.

Then the statement above is the same as $\Delta^{r}H_M(s)=0$ for all $s$ large.

• $H_M(s)$ denotes the dimension of $M_s$ over the field $k$, which is, as I understand a natural number. How can evaluation at a polynomial $p(n)$ with coefficients in $k$ be a natural number? What am I missing? – MIT Jan 4 '15 at 14:07
• @MIT Oh! They mean a polynomial with rational coefficients (that returns integers on the integers. Like $n(n+1)/2$ and things like that). Was that what was causing trouble for understanding the statement? – Pp.. Jan 4 '15 at 14:15
• Yes! I didnt understand it was a polynomial with rational coefficients, they never mentioned it. Thanks though. – MIT Jan 4 '15 at 14:44

It means that $H_M(s) = p(s)$ for some polynomial $p$ of degree $\leq r-1$ for $s$ sufficiently large.