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Let $R$ be a homogeneous noetherian ring with $R_0$ artinian (e.g. $R = k[x_1,\dots, x_n]$), and $M$ a finitely generated graded $R$-module. I want to show that $e(M)$, the multiplicity of $M$, is positive. My notes state this without comment as if it were clear, but I cannot figure it out. Here is some more background behind his definition of the hilbert polynomial:

It is not hard to show the hilbert series, $h_M(t) = \sum_{i\geq 0} H_M(i)$, can be written uniquely as a rational function $q(t)/(1-t)^d$ where $d=\dim M$ (here $H_M(i)$ is the length of $M_i$ over $R_0$). Then $e(M) := q(1)$. In this case (where $R$ is homogeneous noetherian) one can compute the hilbert polynomial explicitly:

$$P_M(x) = \sum_{j=0}^{d-1} (-1)^j e_j \left (\begin{matrix} x+d-1-j \\ d-1-j \end{matrix} \right)$$

where $q(t) = \sum_{j=0}^{s} (-1)^j e_j (1-t)^j$.

I have seen other texts state that the multiplicity (w.r.t a hilbert-samuel function, which is w.r.t an ideal of definition) is a limit of lengths of modules over $R_0$, in which case the answer to my question is clear. I think I should be able to show $e(M)$ is positive as stated above, without appealing to much more general results about the hilbert-samuel functions.

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    $H(M,n)=P_M(n)$ for $n$ large enough. If the dominant coefficient of $P_M$ is negative, then $P_M(n)<0$ for $n$ large enough, a contradiction. (We also have $e_0=Q(1)$.) – 2017-01-17
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    @user26857 Thanks - the first statement you made clears all of the confusion up. But before I posted this question on here I was looking for a while at the fact that $q(1) = e_0$ and could not get anywhere with it. Does that alone tell you it is positive? – 2017-01-17
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    Do you mean how to show that $Q(1)>0$ without considering $Q(1)$ as being related to the Hilbert polynomial, that is, only by using the Hilbert series? – 2017-01-17
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    @user26857 Yes, I was trying to look at the hilbert series and somehow show it by induction or by showing it was some kind of converging limit of positive numbers. – 2017-01-17
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    I can answer this question for $R_0=K$ a field. In this case one can prove it by using a graded Noether normalization $S$ of $R$, and $Q(1)$ equals the rank of $M$ as $S$-module. – 2017-01-18

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