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Let $R$ be an $\mathbb{N}$-graded ring with $R_0$ Artinian and $R = R_0[x_1,\dots,x_r]$, where the degree of $x_i$ is $d_i > 0$. Let $M$ a finitely generated $\mathbb{N}$-graded $R$-module with Hilbert function $H(M,n)$. Then it is known that there exists a unique polynomial $P_M(t)$ such that $H(M,n) = P_M(n)$ for large enough $n$. This result can be found e.g. in Matsumura's Commutative Ring Theory at pages 94-95.

In Bruns&Herzog Cohen-Macaulay Rings, a quasi-polynomial is defined to be a function $f:\mathbb{Z} \rightarrow \mathbb{C}$, such that $f$ is a periodic piecewise polynomial. Then Theorem 4.4.3 reads as follows: enter image description here

Question: I am failing to see in what way the setting of this theorem is a generalization of the setting described in the first paragraph of this question above. This has to be a generalization, since now the statement in (a) involves a quasi-polynomial instead of a polynomial. One possibility that i see is that even though $R$ is concentrated in non-negative degrees, $M$ may now be non-zero in negative degrees as well. But there can be finitely many such negative degrees since $M$ is finitely generated and $R$ is positively graded. So i don't think that the existence of finitely many negative components of $M$ would affect the Hilbert polynomial.

Manos
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Although I don't understand your question very well, let me just say that the Theorem 13.2 in Matsumura says something about the Hilbert series of a finitely generated positively graded $R$-module (this is why $f\in\mathbb Z[t]$). The corollary on the bottom of page 95 gives that the Hilbert function agrees with a polynomial only in the case that the variables have degrees $d_1=\dots=d_r = 1$.

Theorem 4.4.3 in Bruns and Herzog deals with $\mathbb Z$-graded modules and is focused on Hilbert (quasi)polynomial and Hilbert functions rather than on Hilbert series which is treated earlier in Proposition 4.4.1.

It also seems that you want to know why the Hilbert poly is in fact a quasi-polynomial in this case. This follows from 4.4.10 and it is a question related to generating functions and their representation as a rational fraction.

Thomas
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user26857
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  • My question can be alternatively phrased as: which feature of the context in B&H gives rise to a quasi-polynomial instead of a polynomial? As you mention the only difference between the contexts of Matsumura and B&H is that in the latter the module may be $\mathbb{Z}$-graded, while Matsumura considers nonnegatively graded modules. But as i mention in my question, $M$ can only be non-zero at finitely many negative degrees and so this should not affect its Hilbert function for large $n$. So i would still expect the Hilbert series of $M$ to be of polynomial type. Is it clearer now? – Manos Dec 26 '14 at 23:31
  • @Manos As far as I can see Matsumura says something about the Hilbert poly only for the degree one case. – user26857 Dec 26 '14 at 23:33
  • No, i believe i have this distinction clear. Where i got confused is on page 95 in Matsumura. I didn't realize that he talks about "Hilbert Polynomial" in the case where $d_1=\cdots=d_r=1$. But when these degrees are not equal then we have a quasi-polynomial. Correct? – Manos Dec 26 '14 at 23:36
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    @Manos This is the key point. – user26857 Dec 26 '14 at 23:37