Let $X$ an hypersurface in $\mathbb{P}^{n}$ of degree $d$. I would like to prove that the Hilbert polynomial of $X$ is

$\qquad \qquad \qquad \qquad \qquad \qquad p(n)= \begin{pmatrix} n+r \\ n \end{pmatrix} - \begin{pmatrix} n+d-r \\ n \end{pmatrix} $

In the book "Geometry of Algebraic Curves Vol. II", Arbarello, Cornalba, Griffith, p. 7, I read that this result can be proved by taking the cohomology of the following exact sequence

$ \qquad \qquad \qquad \qquad \qquad 0 \longrightarrow \mathcal{O}_{\mathbb{P}^{r}}(n-d) \longrightarrow \mathcal{O}_{\mathbb{P}^{r}}(n) \longrightarrow \mathcal{O}_{\mathbb{P}^{r}}(d) \longrightarrow 0 $

Nevertheless, I am not able to compute $p(n)$.

  • $\begingroup$ I noticed there were some inconsistencies in the use of $d$, $r$ and $n$. I tweaked the question so that it reflects the book you cited - albeit not the notation in the answer below. $\endgroup$
    – Aurelio
    Aug 31, 2020 at 14:39

1 Answer 1


1) Non-cohomological approach:

First, the Hilbert function $h_X:\mathbb{Z} \to \mathbb{Z}$ is defined as

$$h_X: d \mapsto \dim_kS(X)^{(d)}$$

for a projective subscheme $X \subset \mathbb{P}^n,$ where $S(X)$ is the homogeneous coordinate ring of $X$ (which is a graded ring) and $S(X)^{(d)}$ is the homogeneous degree $d$ part of it, which is a finite dimensional vector space over $k.$ Also $h_X(d)=0$ for $d<0.$

Exercise: Show that $h_X(d)=\binom{n+d}{d}$ for $X=\mathbb{P}^n$ and note that $h_X(d)$ is a polynomial in $d$ of degree $n.$

It turns out that for $d \gg 0$ the Hilbert function $h_X(d)$ can be viewed as a polynomial in $d$ called the Hilbert polynomial $p_X(d).$ (Harder exercise!) (Of course, $d \gg 0$ means relatively big enough.) So finding the Hilbert function helps us a lot in order for finding the Hilbert polynomial.

Now, let $X$ be a hypersurface in $\mathbb{P}^n$ given by a degree $r$ homogeneous polynomial $f$ i.e. $X=\text{Proj}k[x_0,\cdots,x_n]/(f).$ You can easily show that

$$\dim_kS(X)^{(d)}=\dim_k k[x_0,\cdots,x_n]^{(d)}-\dim_k k[x_0,\cdots,x_n]^{(d-r)}$$

Therefore, $h_X(d)=\binom{n+d}{d}-\binom{n+d-r}{d-r}$ which is already a polynomial in $d$ so has to be equal to $p_X(d).$

2) Cohomological approach:

The Hilbert polynomial can also be defined as

$$p_X(d)=\chi(\mathcal{O}_X(d))=\sum_{i=0}^n \dim_k (-1)^iH^i(X,\mathcal{O}_X(d))$$

when $d \gg 0$ for a pojective subscheme $X$ of $\mathbb{P}^n.$

As before, assume $X$ is a hypersurface given by a degree $r$ homogeneous polynomial $f$ in $\mathbb{P}^n.$

The correct setting is to first consider the following SES

$$0 \longrightarrow \mathcal{O}_{\mathbb{P}^n}(-r) \stackrel{.f}{\longrightarrow} \mathcal{O}_{\mathbb{P}^n} \longrightarrow \mathcal{O}_X \longrightarrow 0$$

then twist it by $\mathcal{O}(d)$ to get

$$0 \longrightarrow \mathcal{O}_{\mathbb{P}^n}(d-r) \longrightarrow \mathcal{O}_{\mathbb{P}^n}(d) \longrightarrow \mathcal{O}_X(d) \longrightarrow 0$$

Now, since Euler characteristic $\chi$ is additive on SES, we'd have


Using the famous knowledge of sheaf cohomology of $\mathcal{O}_{\mathbb{P}^n}(d)$ you should be able to get the same polynomial for $d \gg 0$ as before.

  • 2
    $\begingroup$ Dear Ehsan M. Kermani, it baffles me that a helpful, thorough answer like this gets no upvotes. +1 from me. One minor comment: in the last sentence, invoking a "vanishing theorem" makes things sound scarier than they need to be --- all you need to know is that $O(k)$ has no higher cohomology when $k$ is positive. (Of course, that is a sort of vanishing theorem, but not of the sort that term usually refers to.) $\endgroup$
    – user64687
    Nov 20, 2013 at 15:41
  • 1
    $\begingroup$ I agree, this is a lovely answer. +1 $\endgroup$
    – Brenin
    Nov 20, 2013 at 16:44
  • $\begingroup$ Thank you dear Asal and Brenin. I've omitted the not so necessary "vanishing theorem" phrase. $\endgroup$ Nov 20, 2013 at 18:12
  • $\begingroup$ Thank you for writing such a good answer $\endgroup$
    – Fq00
    Nov 21, 2013 at 14:07
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    $\begingroup$ If $\chi(\mathcal O_{\mathbb P^n}(d)) = {{n+d}\choose{d}}$, then wouldn't $\chi(\mathcal O_{\mathbb P^n}(d-r))$ equal ${{n+d-r}\choose{d-r}} = {{n+d-r}\choose{n}}$ as opposed to ${n+d-r}\choose{d}$? $\endgroup$
    – Remy
    Nov 16, 2018 at 21:37

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