The Stacks project

Proof. We prove this by induction on $n$. If $n = 0$, then $\mathbf{P}^ n_ k = \mathop{\mathrm{Spec}}(k)$ and any coherent module is $0$-regular and any surjective map is surjective on global sections. Assume $n > 0$. Consider an exact sequence as in the lemma. Let $P' \in \mathbf{Q}[t]$ be the polynomial $P'(t) = P(t) - P(t - 1)$. Let $m'$ be the integer which works for $n - 1$, $r$, and $P'$. By Lemmas 33.35.8 and 33.33.4 we may replace $k$ by a field extension, hence we may assume $k$ is infinite. Apply Lemma 33.35.3 to the coherent sheaf $\mathcal{F}$. The Hilbert polynomial of $\mathcal{F}' = i^*\mathcal{F}$ is $P'$ (see proof of Lemma 33.35.14). Since $i^*$ is right exact we see that $\mathcal{F}'$ is a quotient of $\mathcal{O}_ H^{\oplus r} = i^*\mathcal{O}^{\oplus r}$. Thus the induction hypothesis applies to $\mathcal{F}'$ on $H \cong \mathbf{P}^{n - 1}_ k$ (Lemma 33.35.2). Note that the map $\mathcal{K}(-1) \to \mathcal{K}$ is injective as $\mathcal{K} \subset \mathcal{O}^{\oplus r}$ and has cokernel $i_*\mathcal{H}$ where $\mathcal{H} = i^*\mathcal{K}$. By the snake lemma (Homology, Lemma 12.5.17) we obtain a commutative diagram with exact columns and rows

Thus the induction hypothesis applies to the exact sequence $0 \to \mathcal{H} \to \mathcal{O}_ H^{\oplus r} \to \mathcal{F}' \to 0$ on $H \cong \mathbf{P}^{n - 1}_ k$ (Lemma 33.35.2) and $\mathcal{H}$ is $m'$-regular. Recall that this implies that $\mathcal{H}$ is $d$-regular for all $d \geq m'$ (Lemma 33.35.10).

Let $i \geq 2$ and $d \geq m'$. It follows from the long exact cohomology sequence associated to the left column of the diagram above and the vanishing of $H^{i - 1}(H, \mathcal{H}(d))$ that the map

is injective. As these groups are zero for $d \gg 0$ (Cohomology of Schemes, Lemma 30.14.1) we conclude $H^ i(\mathbf{P}^ n_ k, \mathcal{K}(d))$ are zero for all $d \geq m'$ and $i \geq 2$.

We still have to control $H^1$. First we observe that all the maps

are surjective by the vanishing of $H^1(H, \mathcal{H}(d))$ for $d \geq m'$. Suppose $d > m'$ is such that

is injective. Then $H^0(\mathbf{P}^ n_ k, \mathcal{K}(d)) \to H^0(H, \mathcal{H}(d))$ is surjective. Consider the commutative diagram

By Lemma 33.35.11 we see that the bottom horizontal arrow is surjective. Hence the right vertical arrow is surjective. We conclude that

is injective. By induction we see that

are all injective and we conclude that $H^1(\mathbf{P}^ n_ k, \mathcal{K}(d - 1)) = 0$ because of the eventual vanishing of these groups. Thus the dimensions of the groups $H^1(\mathbf{P}^ n_ k, \mathcal{K}(d))$ for $d \geq m'$ are strictly decreasing until they become zero. It follows that the regularity of $\mathcal{K}$ is bounded by $m' + \dim _ k H^1(\mathbf{P}^ n_ k, \mathcal{K}(m'))$. On the other hand, by the vanishing of the higher cohomology groups we have

Note that the $H^0$ has dimension bounded by the dimension of $H^0(\mathbf{P}^ n_ k, \mathcal{O}^{\oplus r}(m'))$ which is at most $r{n + m' \choose n}$ if $m' > 0$ and zero if not. Finally, the term $\chi (\mathbf{P}^ n_ k, \mathcal{K}(m'))$ is equal to $r{n + m' \choose n} - P(m')$. This gives a bound of the desired type finishing the proof of the lemma. $\square$