# Hilbert polynomial of an hypersurface in projective space

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 $$p(n)$$.

• 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. Aug 31, 2020 at 14:39

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

$$\chi(\mathcal{O}_X(d))=\chi(\mathcal{O}_{\mathbb{P}^n}(d))-\chi(\mathcal{O}_{\mathbb{P}^n}(d-r)).$$

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.

• 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.)
– user64687
Nov 20, 2013 at 15:41
• I agree, this is a lovely answer. +1 Nov 20, 2013 at 16:44
• Thank you dear Asal and Brenin. I've omitted the not so necessary "vanishing theorem" phrase. Nov 20, 2013 at 18:12
• Thank you for writing such a good answer
– Fq00
Nov 21, 2013 at 14:07
• 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 ${n+d-r}\choose{d}$?
– Remy
Nov 16, 2018 at 21:37