Hilbert polynomial and dimension of $M \otimes K(x_1,\dots,x_n)$ Let $K$ a field and  $M$ a finetely generated graded module over $K[x_1,\dots,x_n]$. My goal is to study the dimension of $M \otimes_{K[x_1,\dots,x_n]} K(x_1,\dots,x_n)$ as vector space over $L=K(x_1,\dots,x_n)$
I think I could do it using Hilbert polynomial:  we know that the Hilbert polynomial of $M$ is a polynomial of degree $n-1$, let we write:
$$HP_M(t)=a_{n-1} t^{n-1}+\dots+a_0$$
It seems to me quite intuitively that  $a_{n-1}=\dim_L (M \otimes K(x_1,\dots,x_n))$. 
How could I justify it properly, maybe using some known result?
 A: Here it is a general technique which I find useful in this context.
If you deal with a noetherian ring $A$, you always have a finite graded resolution $ 0 \to A^{n_k} \to \ldots \to A^{n_0} \to M$ with free factors, where some shiftings of graded structure are supposed in $A^{\ell}$ factors. 
Suppose you want to show that two quantities $\lambda(M), \mu(M)$ coincide on finitely generated modules. Furthermore, you know that these quantities are additive on short (graded) exact sequences. Precisely, if $0 \to M' \to M \to M'' \to 0$ is exact, then $\lambda(M) = \lambda(M') + \lambda(M'')$ and similarly for $\mu$. Then $\mu=\lambda$ on fg modules iff $\mu(A^{[d]}) = \lambda(A^{[d]})$ where $A^{[d]}$ is the base ring with shifted grading. Infact, call $Z_r = im ( A^{n_r} \to A^{n_{r-1}} )$. Then you have exact sequences $0 \to Z_{r+1} \to A^{n_r} \to Z_r \to 0$ and $Z_k = 0$, so you can progressively compute $\lambda, \mu$ until $0 \to Z_0 \to A^{n_0} \to M \to 0$, in terms of values on $A^{n_i}$. Finally, note that $\mu(A^{\ell}) = \ell \mu(A)$ by additivity, so we are done.
Now let's show that $\dim_L(M) :=\dim_L(M \otimes L) $ is additive. Take an exact sequence $0 \to M \to N \to P \to 0 $. $L$ is flat over $A$, because it is a localization of $A$. Thus $0\to L \otimes_A M \to L \otimes N \to L \otimes P \to 0$ is exact as vector spaces, and here we can use classical additivity of dimension.
Furthermore, $\dim_L(A^{[d]}) = 1$ disregarding the grading, and $lt(HP_{A^{[d]}}) = lt(t^d HP_A) = lt(HP_A) = 1/(n-1)!$. Thus we conjecture $\dim_L(M) =lt(HP_M)(n-1)! $.
Now I'm not fresh with definitions of HP and I don't get the additivity of its leading term ... But it is equivalent to the original problem. Infact if the claim is true, then the leading term is a function ($\dim_L$) that i have shown to be additive.
