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?

• @user26857 It's a graded module, of course – user84976 Jan 15 '18 at 16:50
• Actually the dimension is $(n-1)!a_{n-1}$, that is, the multiplicity of $M$. – user26857 Jan 15 '18 at 20:24
• @user26857 That seems what I am looking for. Is there a good reference for this statement? – user84976 Jan 16 '18 at 9:10
• The dimension you are looking for is the rank of $M$ (in the sense of Definition 1.4.2 from Bruns and Herzog). The leading coefficient of the Hilbert poly is $e(M)/(d-1)!$, where $d=\dim M$ and $e(M)$ the multiplicity of $M$. On the other side $e(M)=Q_M(1)$ (see Proposition 4.1.9 from the same book). But $Q_M(1)$ is the rank of $M$ (see Lemma 4.1.13 and Corollary 4.1.14). The last claim isn't explicit in the book, but it's easy to find the rank of a module via resolutions. – user26857 Jan 18 '18 at 10:06

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.

• I agree with yout and I have already done this reasoning also thanks to user26857 comment, but I'd like almost all find a clear reference for the statement. Anyway the bounty is your – user84976 Jan 25 '18 at 11:16