The Stacks project

Proof. We have $\mathbf{P}^ n_ k = \text{Proj}(k[T_0, \ldots , T_ n])$. The section $s$ corresponds to a homogeneous form in $T_0, \ldots , T_ n$ of degree $1$, see Cohomology of Schemes, Section 30.8. Say $s = \sum a_ i T_ i$. Constructions, Lemma 27.13.7 gives that $H = \text{Proj}(k[T_0, \ldots , T_ n]/I)$ for the graded ideal $I$ defined by setting $I_ d$ equal to the kernel of the map $\Gamma (\mathbf{P}^ n_ k, \mathcal{O}(d)) \to \Gamma (H, i^*\mathcal{O}(d))$. By our construction of $Z(s)$ in Divisors, Definition 31.14.8 we see that on $D_{+}(T_ j)$ the ideal of $H$ is generated by $\sum a_ i T_ i/T_ j$ in the polynomial ring $k[T_0/T_ j, \ldots , T_ n/T_ j]$. Thus it is clear that $I$ is the ideal generated by $\sum a_ i T_ i$. Note that

as graded rings. For example, if $a_ n \not= 0$, then mapping $S_ i$ equal to the class of $T_ i$ works. We obtain the desired isomorphism by functoriality of $\text{Proj}$. Equality of twists of structure sheaves follows for example from Constructions, Lemma 27.11.5. $\square$

Proof. The map $\mathcal{F}(-1) \to \mathcal{F}$ comes from Constructions, Equation (27.10.1.2) with $n = 1$, $m = -1$ and the section $s$ of $\mathcal{O}(1)$. Let's work out what this map looks like if we restrict it to $D_{+}(T_0)$. Write $D_{+}(T_0) = \mathop{\mathrm{Spec}}(k[x_1, \ldots , x_ n])$ with $x_ i = T_ i/T_0$. Identify $\mathcal{O}(1)|_{D_{+}(T_0)}$ with $\mathcal{O}$ using the section $T_0$. Hence if $s = \sum a_ iT_ i$ then $s|_{D_{+}(T_0)} = a_0 + \sum a_ ix_ i$ with the identification chosen above. Furthermore, suppose $\mathcal{F}|_{D_{+}(T_0)}$ corresponds to the finite $k[x_1, \ldots , x_ n]$-module $M$. Via the identification $\mathcal{F}(-1) = \mathcal{F} \otimes \mathcal{O}(-1)$ and our chosen trivialization of $\mathcal{O}(1)$ we see that $\mathcal{F}(-1)$ corresponds to $M$ as well. Thus restricting $\mathcal{F}(-1) \to \mathcal{F}$ to $D_{+}(T_0)$ gives the map

To see that the arrow is injective, it suffices to pick $a_0 + \sum a_ ix_ i$ outside any of the associated primes of $M$, see Algebra, Lemma 10.63.9. By Algebra, Lemma 10.63.5 the set $\text{Ass}(M)$ of associated primes of $M$ is finite. Note that for $\mathfrak p \in \text{Ass}(M)$ the intersection $\mathfrak p \cap \{ a_0 + \sum a_ i x_ i\} $ is a proper $k$-subvector space. We conclude that there is a finite family of proper sub vector spaces $V_1, \ldots , V_ m \subset \Gamma (\mathbf{P}^ n_ k, \mathcal{O}(1))$ such that if we take $s$ outside of $\bigcup V_ i$, then multiplication by $s$ is injective over $D_{+}(T_0)$. Similarly for the restriction to $D_{+}(T_ j)$ for $j = 1, \ldots , n$. Since $k$ is infinite, a finite union of proper sub vector spaces is never equal to the whole space, hence we may choose $s$ such that the map is injective. The cokernel of $\mathcal{F}(-1) \to \mathcal{F}$ is annihilated by $\mathop{\mathrm{Im}}(s : \mathcal{O}(-1) \to \mathcal{O})$ which is the ideal sheaf of $H$ by Divisors, Definition 31.14.8. Hence we obtain $\mathcal{G}$ on $H$ using Cohomology of Schemes, Lemma 30.9.8. $\square$

Proof. This is true because

by flat base change, see Cohomology of Schemes, Lemma 30.5.2. $\square$

Proof. Recall that $H^ i(\mathbf{P}^ n_ k, i_*\mathcal{G}) = H^ i(H, \mathcal{G})$ by Cohomology of Schemes, Lemma 30.2.4. Hence we see that for $i \geq 1$ we get

by Remark 33.35.5. The lemma follows. $\square$

Proof. We prove this by induction on $n$. If $n = 0$ every sheaf is $m$-regular for all $m$ and there is nothing to prove. By Lemma 33.35.8 we may replace $k$ by an infinite overfield and assume $k$ is infinite. Thus we may apply Lemma 33.35.3. By Lemma 33.35.9 we know that $\mathcal{G}$ is $m$-regular. By induction on $n$ we see that $\mathcal{G}$ is $(m + 1)$-regular. Considering the long exact cohomology sequence associated to the sequence

as in Remark 33.35.5 the reader easily deduces for $i \geq 1$ the vanishing of $H^ i(\mathbf{P}^ n_ k, \mathcal{F}(m + 1 - i))$ from the (known) vanishing of $H^ i(\mathbf{P}^ n_ k, \mathcal{F}(m - i))$ and $H^ i(\mathbf{P}^ n_ k, \mathcal{G}(m + 1 - i))$. $\square$

Proof. Let $k'/k$ be an extension of fields. Let $\mathcal{F}'$ be as in Lemma 33.35.8. By Cohomology of Schemes, Lemma 30.5.2 the base change of the linear map of the lemma to $k'$ is the same linear map for the sheaf $\mathcal{F}'$. Since $k \to k'$ is faithfully flat it suffices to prove the lemma over $k'$, i.e., we may assume $k$ is infinite.

Assume $k$ is infinite. We prove the lemma by induction on $n$. The case $n = 0$ is trivial as $\mathcal{O}(1) \cong \mathcal{O}$ is generated by $T_0$. For $n > 0$ apply Lemma 33.35.3 and tensor the sequence by $\mathcal{O}(m + 1)$ to get

see Remark 33.35.5. Let $t \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(m + 1))$. By induction the image $\overline{t} \in H^0(H, \mathcal{G}(m + 1))$ is the image of $\sum g_ i \otimes \overline{s}_ i$ with $\overline{s}_ i \in \Gamma (H, \mathcal{O}(1))$ and $g_ i \in H^0(H, \mathcal{G}(m))$. Since $\mathcal{F}$ is $m$-regular we have $H^1(\mathbf{P}^ n_ k, \mathcal{F}(m - 1)) = 0$, hence long exact cohomology sequence associated to the short exact sequence

shows we can lift $g_ i$ to $f_ i \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(m))$. We can also lift $\overline{s}_ i$ to $s_ i \in H^0(\mathbf{P}^ n_ k, \mathcal{O}(1))$ (see proof of Lemma 33.35.2 for example). After subtracting the image of $\sum f_ i \otimes s_ i$ from $t$ we see that we may assume $\overline{t} = 0$. But this exactly means that $t$ is the image of $f \otimes s$ for some $f \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(m))$ as desired. $\square$

Proof. For all $d \gg 0$ the sheaf $\mathcal{F}(d)$ is globally generated. This follows for example from the first part of Cohomology of Schemes, Lemma 30.14.1. Pick $d \geq m$ such that $\mathcal{F}(d)$ is globally generated. Choose a basis $f_1, \ldots , f_ r \in H^0(\mathbf{P}^ n_ k, \mathcal{F})$. By Lemma 33.35.11 every element $f \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(d))$ can be written as $f = \sum P_ if_ i$ for some $P_ i \in k[T_0, \ldots , T_ n]$ homogeneous of degree $d - m$. Since the sections $f$ generate $\mathcal{F}(d)$ it follows that the sections $f_ i$ generate $\mathcal{F}(m)$. $\square$

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

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). $\square$

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

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. $\square$