# Interpretation of the numerator of the Hilbert series?

Let $R$ be a finitely generated graded ring over a field $k$. Let $R_\ell$ be the degree-$\ell$ homogeneous component of $R$.

By the Noether normalization theorem, $R$ is finite over a graded polynomial algebra $S = k[\theta_1,\dots,\theta_n]$, where the $\theta_i$ are a homogeneous system of parameters. Let $d_1,\dots,d_n$ be the degrees of the $\theta_i$.

Then the Hilbert series of $R$ has the form

$$\sum_{\ell \geq 0} (\dim_kR_\ell) t^\ell = \frac{\text{polynomial}}{\prod_1^n \left(1-t^{d_i}\right)}.$$

One proof is to observe that $R$ has a finite graded free resolution (of length $\leq n$) by the Hilbert syzygy theorem, and then to explicitly write down the Hilbert series using the resolution.

My question concerns the polynomial in the numerator; in particular its value at $t=1$. Call this polynomial $p(t)$. If $R$ happens to be a Cohen-Macaulay ring, then $R$ is already free as an $S$-module, and $p(t)$ explicitly gives the degrees of the generators for $R$ as an $S$-module. Then $p(1)$ is just the rank of $R$ as $S$-module.

What happens if $R$ is not Cohen-Macaulay? The term-by-term interpretation of $p(t)$ is now too much to hope for since coefficients may be negative, but:

Is $p(1)$ still the rank of something? In particular, if $K$ is $S$'s fraction field, is it the dimension of $R\otimes_S K$ over $K$?

(As often happens at this site, as I wrote out the question I began to think more clearly about it, and I think the answer is yes, as follows. Consider the graded free resolution of $R$ over $S$. Each module in the resolution has an analogous number $p_j(1)$ where $p_j$ is the numerator of the Hilbert series of the $j$th module in the resolution, and we should have

$$p(1) = \sum_{j\geq 0} (-1)^jp_j(1)$$

because (i) in each degree we have an exact sequence of finite dimensional $k$-vector spaces in which dimension is additive in short exact sequences, so the analogous relation holds for the entire Hilbert series; (ii) everything is over a common denominator so the analogous relation holds in the numerators; (iii) so it certainly continues to hold when you specialize to $t=1$.

These numbers $p_j(1)$ are just the ranks of the modules in the resolution, because they are all free. Tensoring the resolution by $K$ preserves exactness because $K$ is flat over $S$ [localization is exact], but now everything is a vector space over $K$ and all the $p_j(1)$'s can be interpreted as vector space dimensions of their corresponding modules. Since vector space dimension is additive in short exact sequences, the relation $p(1) = \sum (-1)^jp_j(1)$ now implies that $p(1)$ coincides with the $K$-dimension of $R\otimes_S K$. Do you buy it?)

• You always get that $p(1)$ is the rank of $R$ as an $S$-module (you almost proved it!); see e.g. Bruns and Herzog, Lemma 4.1.13. – user26857 Dec 22 '17 at 16:48
• @user26857 - By rank of $R$ as $S$-module you mean dimension of $R\otimes_S K$ as $K=\operatorname{Frac} S$-vector space, right? In which case, I now think I did prove it! The argument in B-H 4.1.13 is more concise but I think the substance is the same. – Ben Blum-Smith Dec 22 '17 at 21:22
• Yes, that's what I mean. You can also use Corollary 1.4.6 (same book). – user26857 Dec 22 '17 at 21:25
• Thanks! (The relevant result is corollary 1.4.5 in my copy.) – Ben Blum-Smith Dec 26 '17 at 2:50
• It seems you use the 1993 edition, that's why. (Be careful, that edition contains many mistakes.) – user26857 Dec 26 '17 at 8:26

## 1 Answer

The answer is yes. The proof in the OP is correct.

In comments, @user26857 points to Winfried Bruns and Jurgen Herzog's book Cohen-Macaulay Rings, where various forms of the result also appear. The special case where all the $d$'s equal $1$ appears as 4.1.13 in that book. In my edition, the underlying result on the rank of the module appears as 1.4.5. (@user26857 cites 1.4.6 so I assume it appears differently in different editions.) In both cases, the substance of the proof is essentially that given in the OP.