# is the hilbert polynomial integer-valued everywhere?

Let $R$ be an $\mathbb{N}$-graded Noetherian ring, finitely generated over $R_0$ with $R_0$ local Artinian. Let $M$ be a finite $R$-module of Krull dimension $d$. It is known that the Hilbert function $H(M,n) = \operatorname{length} (M_n)$ coincides with a polynomial $h(n)$ of degree $d-1$ for large values of $n$.

Question: Certainly, $h(n)$ is integer valued for large values of $n$, but how about for small values of $n$?

Remark: The discussion in Bruns and Herzog, Cohen-Macaulay Rings, pages 149-150, seems to imply that $h(n) \in \mathbb{Z}$ for small values of $n$ (see in particular remark 4.1.6).

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It's a general principle that a polynomial $f(x)$ in $\mathbb Q(x)$ for which $f(n)$ is an integer for large values of $n$ is in fact integer valued on all integers $n$.
One way to see this is as follows: By continuity, if $x$ is a $p$-adic integer for any prime $p$, then $f(x)$ is again a $p$-adic integer. (Polynomials are continuous, and the set of integers greater than any given $n_0$ is dense in $\mathbb Z_p$.) Thus if $n$ is an integer, $f(n)$ is a rational number that is also a $p$-adic integer for every prime $p$, and hence is an integer.
In the case of Hilbert polynomials, one can also interpret the value $h(n)$ as an Euler characteristic for every value of $n$, and so see that it is integer valued directly.
Thanks for your answer. $h(n)$ is an euler characteristic of which complex? –  Manos Mar 31 '14 at 1:48
Dear Manos, One can form $X :=$ Proj $R$, and then localize $M$ to obtain a coherent sheaf $\mathcal M$ on $X$. We then have that $h(n) = \sum_i (-1)^i \operatorname{length}\Bigl(H^i\bigl(X,\mathcal M(n)\bigr)\Bigr).$ Regards, –  Matt E Mar 31 '14 at 3:04