Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 113 (Algebraic Geometry)

The points of the spectrum of a commutative ring $A$ are its prime ideals. Its Zariski topology has closed sets $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$ and basis $D(f)=\{\mathfrak{p}:f\notin \mathfrak{p}\}$. The structure sheaf of a scheme can be defined as follows: a section over $U$ assigns an element of $A_{\mathfrak{q}}$ to every $\mathfrak{q}\in U$, and locally that assignment is represented by one fraction $a/s$, with $s$ outside every prime in the neighbourhood. Restrictions forget values. Local representability and equality can be checked on an open cover, so these assignments satisfy the sheaf gluing axiom. On $D(f)$ the sections identify with the localization of a ring $A_f$.

Evaluation at $\mathfrak{p}$ induces the ring homomorphism ${\mathcal{O}}_{X,\mathfrak{p}}\to A_{\mathfrak{p}}$. It is surjective because $a/s\in A_{\mathfrak{p}}$ is represented by a section on $D(s)$. For injectivity, represent a germ by $a/s$ on a neighbourhood of $\mathfrak{p}$. If its value in $A_{\mathfrak{p}}$ vanishes, there is $t\notin \mathfrak{p}$ with $ta=0$. The section is then zero after restricting to $D(st)$ inside that neighbourhood. Its germ is zero. Thus

Put $A=k[t_1,t_2]$. The punctured affine plane is covered by the principal open subschemes $D(t_1)$ and $D(t_2)$. Since $A$ is an integral domain, restriction identifies its global sections withThis intersection is $A$: if $a/t_1^r=b/t_2^s$, then $t_2^{s}a=t_1^{r}b$, and unique factorization forces $t_1^r$ to divide $a$. Therefore the inclusion $j:X\hookrightarrow \mathrm{Spec}A$ induces the identity on global regular functions.

If $X$ were an affine scheme, affine scheme reconstruction from global sections would make $j$ an isomorphism of schemes, because its global-section map is an isomorphism. But its image omits the closed point $(t_1,t_2)$. The punctured affine plane is therefore not affine. This is a useful example of global regular functions failing to distinguish a nonaffine open subscheme from its affine ambient space.

An isomorphism of schemes over $\mathbb{R}$ induces a bijection on rational points. A real point of this projective plane curve would be represented by a nonzero real triple with $t_0^2+t_1^2+t_2^2=0$. Each summand is nonnegative, so every coordinate would vanish. Thus $X(\mathbb{R})=\varnothing$, whereas $[1:0]\in {\mathbb{P}}^1(\mathbb{R})$. The two real schemes are not isomorphic. The obstruction is the real conic without real rational points; nonsingularity alone does not make a real conic a projective line over its ground field.

A fibre product of schemes $Z=X{\times }_{S}Y$ has projection morphisms to $X$ and $Y$ with equal composites to $S$, and for every scheme $T$ the induced map is a bijectionThis universal property determines $Z$ uniquely up to unique isomorphism. On affine charts over $\mathrm{Spec}R$, takeThe tensor product of commutative algebras has the required universal property for ring maps. Localization identifies these constructions on principal-open overlaps. Cover the base and its two inverse images by affine charts and glue the resulting affine products along those overlaps; their local universal properties give the global one.

For a counterexample to preservation of irreducibility, take $X=Y=\mathrm{Spec}\mathbb{C}$ and $S=\mathrm{Spec}\mathbb{R}$ with their usual structure morphisms. All three are integral schemes, but the Chinese remainder theorem givesThe product scheme is the disjoint union of two points, so it is not irreducible.

Work first over an affine neighbourhood $\mathrm{Spec}A$ of $y=\mathfrak{p}$, with an affine chart $\mathrm{Spec}B$ in its inverse image. The scheme-theoretic fibre on this chart has ringwhere $\kappa (\mathfrak{p})$ is the residue field. The prime ideal correspondence for localization and the correspondence for a quotient identify its points with primes $\mathfrak{q}\subset B$ containing $\mathfrak{p}B$ and disjoint from $S$. These two conditions are exactly $\mathfrak{q}\cap A=\mathfrak{p}$.

Under this bijection, each principal open $D(b\otimes 1)$ corresponds to $D(b)\cap f^{-1}\{y\}$. These are bases for the two topologies: every element of the fibre ring is represented by $b/s$, with the image of $s$ invertible. Thus the bijection is a homeomorphism. The identifications agree on overlapping charts and glue, provingThe scheme-theoretic fibre may nevertheless carry nilpotents, so this topological statement does not identify its structure sheaf with a naive restriction of ${\mathcal{O}}_X$.

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.

A flasque sheaf has surjective restriction maps $\mathcal{F}(V)\to \mathcal{F}(U)$ for every pair of opens $U\subseteq V$. To prove the claim for an injective sheaf of modules, let $j_U:U\hookrightarrow X$ and $j_V:V\hookrightarrow X$. The natural mapis a monomorphism: its stalks are either the identity on ${\mathcal{O}}_{X,x}$, the map from zero to that stalk, or the zero-to-zero map. Here $j_{!}$ is extension by zero for module sheaves.

The extension-by-zero adjunction identifiesThe injective object property extends every morphism from $(j_U{)}_{!}{\mathcal{O}}_U$ to one from $(j_V{)}_{!}{\mathcal{O}}_V$. Under the displayed identification this is exactly surjectivity of $I(V)\to I(U)$. Injective module sheaves are therefore flasque. This argument works on an arbitrary ringed space, without Noetherian or separation assumptions.

For the closed inclusion $f$, the direct image sheaf has stalk ${\mathcal{F}}_x$ at $x\in X$ and zero stalk at points outside $X$: outside the closed set there is an open neighbourhood disjoint from $X$. Hence $f_{\ast }$ is exact, since exact sequences of sheaves are detected on stalks.

Take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$. Direct image preserves flasqueness, because restrictions are the restrictions on inverse-image opens. Exactness makes $f_{\ast }{\mathcal{I}}^{\bullet }$ a flasque resolution of $f_{\ast }\mathcal{F}$. MoreoverThe complexes computing sheaf cohomology are identical. ThereforeThis is sheaf cohomology under a closed inclusion; the closedness hypothesis is used in the exactness of direct image.

For a point $x\in X$ and an abelian group $A$, the skyscraper sheaf $i_{\ast }A$, with $i:\{x\}\hookrightarrow X$, isRestrictions are identities or the map to zero, so it is a flasque sheaf. The global sections equal $A$ and higher cohomology vanishes:No closed-point assumption is needed for this calculation of skyscraper sheaf cohomology. On an arbitrary topological space, its nonzero stalks are at points of $\overline{\{x\}}$ when $A\ne 0$; only for a closed point is the nonzero stalk confined to $x$.

The direct image of a coherent sheaf under a closed immersion puts $\mathcal{G}=f_{\ast }\mathcal{F}$ in the category of coherent sheaves on ${\mathbb{P}}_k^n$. By sheaf cohomology under a closed inclusion and the supplied compatibility of twisting with direct image,Here is a proof of the required Serre vanishing on projective space. A coherent sheaf on ${\mathbb{P}}_k^n$ is the sheaf associated with a graded module for a finite graded $k[t_0,\dots ,t_n]$-module. Equivalently, it has a presentation by finite sums of twisting sheaves. Use a finite twisting resolution of a coherent sheaf on projective space: resolve the graded module by a finite graded free resolution, using the Hilbert syzygy theorem, and sheafify; exactness of localization preserves the resolution. Its terms are finite sums of $\mathcal{O}(a)$.

Choose $d$ sufficiently large that all twists $a+d$ occurring in these finitely many terms are nonnegative. The cohomology of twisting sheaves on projective space then vanishes in every positive degree for every resolution term. In a short exact sequence $0\to {\mathcal{K}}_{j+1}\to {\mathcal{E}}_j\to {\mathcal{K}}_j\to 0$, with ${\mathcal{K}}_0=\mathcal{G}$, the long exact sequence in sheaf cohomology identifies $H^p({\mathcal{K}}_j(d))$ with $H^{p+1}({\mathcal{K}}_{j+1}(d))$ for $p>0$. Iterating to the final acyclic term provesThe standard graded-module description used here is given in Stacks Project, Section 30.15; the vanishing follows from the displayed finite-resolution argument.

Choose the rational point $x=[1:0:\cdots :0]$ and the coherent sheaf $\mathcal{F}=i_{\ast }k$ supported there. By skyscraper sheaf cohomology, $H^0(X,\mathcal{F})=k$.

Its dual of a sheaf ${\mathcal{F}}^{\vee }=\mathcal{H}om(\mathcal{F},{\mathcal{O}}_X)$ is zero. Away from $x$ its source is zero. At $x$, a homomorphism $k\to {\mathcal{O}}_{X,x}$ must send one to an element annihilated by the maximal ideal. Since $n>0$, that ideal contains a nonzero coordinate parameter; the local ring is an integral domain, so the image must vanish. This proves that every local homomorphism vanishes. ConsequentlyThere is no contradiction with Serre duality: the ordinary sheaf dual suffices for locally free sheaves, but general coherent sheaves require an Ext functor. This example illustrates failure of ordinary sheaf-dual Serre duality for a skyscraper sheaf.

Eliminate $t_3$ using the linear equation. The homogeneous coordinate ring becomesThe cubic is a nonzero element of a polynomial integral domain, hence a non-zero-divisor. It gives the exact sequence of graded modulesThus $\dim B_m=\left(\genfrac{}{}{0ex}{}{m+2}{2}\right)-\left(\genfrac{}{}{0ex}{}{m-1}{2}\right)$ for $m\ge 3$. Equivalently, use the Hilbert series $(1-z^3)/(1-z{)}^3$, or the sheafified exact sequence on ${\mathbb{P}}^2$. The Hilbert polynomial of the plane cubic isIts degree is three and its arithmetic genus is one, although the curve is singular: its affine equation near $[1:0:0]$ is the cusp $t_2^2+t_1^3=0$. The Hilbert polynomial records the arithmetic genus, rather than the genus of the normalization.