Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 113 / 3

The degree-one generation condition for Proj ensures that the opens $D_{+}(f)$ for $f\in S_1$ cover $X$. Put $B=S_f$ and $R=B_0$. Multiplication by $f^n$ is an $R$-linear bijection $R\to B_n$ for every integer $n$, because $f$ is an invertible degree-one element. The convention $S(n{)}_d=S_{n+d}$ therefore identifies $\widetilde{S(n)}$ on this chart with the free rank-one $R$-module $B_n$, generated by $f^n$. Hence every twisting sheaf ${\mathcal{O}}_X(n)$ is invertible, including negative twists.

For the tensor product of sheaves, consider the natural mapIt is an isomorphism by degree-one localization of a graded module. Explicitly, a homogeneous tensor $m_a\otimes n_{-a}$ of total degree zero is represented on the left by $f^{-a}{m}_a\otimes f^a{n}_{-a}$. Multiplying a tensor factor by a homogeneous element of $B$ gives the same result after using the tensor relation, since that element is a degree-zero coefficient times a power of $f$. This defines the inverse. Since localization commutes with tensor products, the target is $((M{\otimes }_{S}N{)}_f{)}_0$. These natural chart maps agree on overlaps, soThe degree-one hypothesis matters: on general Proj constructions, twisting sheaves need not be line bundles. The local trivializations are also recorded in Stacks Project, Section 27.10.

Take the hyperplane divisor $D=(t_0=0)$. On $U_i=D_{+}(t_i)$ its local equation is $g_i=t_0/t_i$, a non-zero-divisor; on $U_0$ it is the unit one. Thus it is an effective Cartier divisor. On overlaps, $g_i/g_j=t_j/t_i$ is a unit.

The divisor line bundle ${\mathcal{O}}_X(D)$ is locally generated by $1/g_i=t_i/t_0$. Map this generator to the local generator $t_i$ of the twisting sheaf on projective space ${\mathcal{O}}_X(1)$. On overlaps both generators transform by $t_i/t_j$, so the maps glue and giveFor $n=0$ the hyperplane is empty and this is the zero Cartier divisor; the conclusion still holds.

The first implication is false. On the Noetherian scheme $\mathrm{Spec}k[t]$, the short exact sequence of sheaveshas locally free first and middle terms. The quotient is not locally free: at $(t)$ it is a nonzero module annihilated by $t$, whereas a free module over the integral domain $k[t{]}_{(t)}$ has no such element.

The second implication is true. Near any point trivialize $\mathcal{F}$ and $\mathcal{H}$. Choose a finite local basis of $\mathcal{H}$. Surjectivity as a sheaf gives local lifts of its basis sections; shrink to the intersection of their finitely many neighbourhoods. The lifts define a splitting $\mathcal{H}\to \mathcal{G}$ on that neighbourhood. Thus $\mathcal{G}\cong \mathcal{F}\oplus \mathcal{H}$ there and is locally free. This is local splitting when a quotient sheaf is locally free; no global splitting is asserted.