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

is a polynomial.

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). $\square$

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