### 33.35.13 Hilbert polynomials

The following lemma will be made obsolete by the more general Lemma 33.45.1.

Lemma 33.35.14. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$. The function

$d \longmapsto \chi (\mathbf{P}^ n_ k, \mathcal{F}(d))$

is a polynomial.

Proof. We prove this by induction on $n$. If $n = 0$, then $\mathbf{P}^ n_ k = \mathop{\mathrm{Spec}}(k)$ and $\mathcal{F}(d) = \mathcal{F}$. Hence in this case the function is constant, i.e., a polynomial of degree $0$. Assume $n > 0$. By Lemma 33.33.4 we may assume $k$ is infinite. Apply Lemma 33.35.3. Applying Lemma 33.33.2 to the twisted sequences $0 \to \mathcal{F}(d - 1) \to \mathcal{F}(d) \to i_*\mathcal{G}(d) \to 0$ we obtain

$\chi (\mathbf{P}^ n_ k, \mathcal{F}(d)) - \chi (\mathbf{P}^ n_ k, \mathcal{F}(d - 1)) = \chi (H, \mathcal{G}(d))$

See Remark 33.35.5. Since $H \cong \mathbf{P}^{n - 1}_ k$ by induction the right hand side is a polynomial. The lemma is finished by noting that any function $f : \mathbf{Z} \to \mathbf{Z}$ with the property that the map $d \mapsto f(d) - f(d - 1)$ is a polynomial, is itself a polynomial. We omit the proof of this fact (hint: compare with Algebra, Lemma 10.58.5). $\square$

Definition 33.35.15. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$. The function $d \mapsto \chi (\mathbf{P}^ n_ k, \mathcal{F}(d))$ is called the Hilbert polynomial of $\mathcal{F}$.

The Hilbert polynomial has coefficients in $\mathbf{Q}$ and not in general in $\mathbf{Z}$. For example the Hilbert polynomial of $\mathcal{O}_{\mathbf{P}^ n_ k}$ is

$d \longmapsto {d + n \choose n} = \frac{d^ n}{n!} + \ldots$

This follows from the following lemma and the fact that

$H^0(\mathbf{P}^ n_ k, \mathcal{O}_{\mathbf{P}^ n_ k}(d)) = k[T_0, \ldots , T_ n]_ d$

(degree $d$ part) whose dimension over $k$ is ${d + n \choose n}$.

Lemma 33.35.16. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$ with Hilbert polynomial $P \in \mathbf{Q}[t]$. Then

$P(d) = \dim _ k H^0(\mathbf{P}^ n_ k, \mathcal{F}(d))$

for all $d \gg 0$.

Proof. This follows from the vanishing of cohomology of high enough twists of $\mathcal{F}$. See Cohomology of Schemes, Lemma 30.14.1. $\square$

Comment #3006 by Noah Olander on

I think in the proof of Lemma 32.34.13 you're using the projection formula without saying so. We know that H^i(H, G) = H^i (X, i_G) but to conclude that H^i (H, G(d)) = H^i (X, (i_G)(d)) maybe you should say that (i_G)(d) = i_(G(d)) by the projection formula.

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