Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 16 3 Solution Created 2026-10-03 Updated 2026-10-06
For , the Čech cochain complex has groupsTerms in cancel in pairs, and Čech cohomology is . A degree-zero cocycle is exactly a family of compatible local sections. The sheaf gluing axiom gives their unique global section, proving . If the cover has affine finite intersections, a quasi-coherent sheaf has no higher cohomology on those intersections by vanishing of quasi-coherent cohomology on an affine scheme. The acyclic cover theorem then identifies all Čech groups with sheaf cohomology. In particular, a finite affine open cover of a separated variety has this property.
For the sheaf of units of the structure sheaf, a multiplicative degree-one cocycle is a family satisfying , with . It glues trivial rank-one free modules into an invertible sheaf. Changing the local frames multiplies by a coboundary, and two sets of transition data give isomorphic line bundles precisely when their cocycles differ this way. Tensoring line bundles multiplies their cocycles. Thus line bundles trivialized by an open cover give the group isomorphismFor the remaining arguments, work over the algebraically closed ground field. On the irreducible variety , put , a quotient of sheaves of abelian groups. A global section of is locally represented by rational functions whose ratios are regular units. The corresponding unit cocycle defines an invertible sheaf. A single global rational function has trivial cocycle. Conversely, every line bundle has a nonzero rational section: choose a nonzero vector in its one-dimensional fibre at the generic point and express it in local frames. This supplies such local . If the associated line bundle is trivial, changing frames makes all restrictions of one rational function. Thereforeis exact. This is the Cartier-divisor description of the Picard group. The sheaf of nonzero rational functions on an irreducible variety is flasque, since all restrictions between nonempty open sets are the identity on . Apply the long exact sequence in sheaf cohomology to . Since , its connecting map has exactly the cokernel just computed, provingFinally the Segre description of a smooth quadric surface identifies with . The two rulings of a smooth quadric surface have classes generating . For completeness, remove one line in each ruling: the remaining chart is the affine plane, with factorial coordinate ring and trivial divisor class group. The localization sequence for the divisor class group makes generators, and their degrees on the two ruling lines prove independence. The hyperplane class, and hence the conic , has bidegree . By Picard-group localization on a smooth variety, the Picard group of a smooth affine quadric surface isExplicitly, the restriction of is nontrivial: if it were trivial on , its rational trivialization would have divisor supported on , forcing to be an integer multiple of , which is impossible.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 20 1 b Solution Created 2026-10-03 Updated 2026-10-06
The affine-target adjunction for schemes gives the natural bijectionA morphism of schemes gives the homomorphism on global sections induced by , using .
Conversely, let be a ring homomorphism. Choose an affine open subscheme cover of . Restriction of global sections gives homomorphisms , hence morphisms of schemes . On any affine open subscheme , both restrictions correspond to the same homomorphism . They therefore agree on . Such affine opens cover the overlap, so the glue uniquely to a morphism of schemes .
The two constructions are inverse: the first recovers on each , hence on all of , and the second recovers every restriction of a given . This proof needs neither affineness nor quasi-compactness of .
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 20 3 c Solution Created 2026-10-03 Updated 2026-10-06
Let . Its degree is five. Because the linear span of has dimension at least three, restrictions of linear forms supply at least four independent global sections, soIf , the Riemann-Roch theorem gives , and therefore immediately.
If instead is special, the Clifford inequality for curves givesso its integer dimension is at most three, a contradiction. Here is a short proof of the required Clifford inequality for curves, so the bound need not be assumed. Put and . At a smooth geometric point, choose bases of these two section spaces with strictly increasing orders of vanishing and . Such bases exist because each successive order-of-vanishing quotient has dimension at most one. The products of the first basis with the first element of the second basis, followed by products of the last element of the first basis with the remaining elements of the second, have ordersThey are linearly independent global sections of , proving the product dimension bound for sections on a curve. Thus . Combining this with Riemann-Roch theorem, , gives , which is exactly the displayed Clifford inequality for curves.
The special case is impossible, and henceThis is the genus bound for a nonplanar degree-five curve. The geometric argument can be carried out after extension to an algebraic closure, which preserves these section dimensions.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 20 5 a Solution Created 2026-10-03 Updated 2026-10-06
A line bundle is a globally generated line bundle when the evaluation mapis surjective. Equivalently, its global sections span each fibre of the line bundle; locally at every point, some section is a generator.
Choose a finite generating family . On the open set where generates , the ratios are regular functions, giving a morphism of schemes into the standard chart of projective space. The ratios obey the usual transition rules on overlaps, so these chart maps glue toThe tuple is computed using any local trivialization of ; changing that trivialization multiplies all entries by the same invertible function. The construction has and pulls back the coordinate sections to the chosen .
There is a finiteness qualification for an arbitrary : global generation alone need not provide such a finite family. It does if is quasi-compact, since the open sets on which individual sections generate have a finite subcover. Without that hypothesis, take and let restrict to on each component. This line bundle is globally generated, but any global sections have a common zero on a component with . Thus this is a globally generated line bundle without finite generators. This is finite global generation on a quasi-compact scheme. The finite-family construction is automatic in the projective case asked next.
For projective nonsingular , put . The length-two criterion for a very ample linear system says that is a closed immersion precisely when, after extending to an algebraic closure, separates distinct points and tangent directions. Equivalently, the evaluationis surjective for every length-two geometric closed subscheme . Two distinct points give point separation; a nonreduced length-two subscheme supported at one point gives separation of a direction in the Zariski tangent space. For the complete space , this says exactly that is a very ample line bundle. For a chosen smaller , it is the chosen linear system of divisors which must be a very ample linear system; mere global generation is insufficient.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 113 1 ii Solution Created 2026-10-03 Updated 2026-10-06
Put . The punctured affine plane is covered by the principal open subschemes and . Since is an integral domain, restriction identifies its global sections withThis intersection is : if , then , and unique factorization forces to divide . Therefore the inclusion induces the identity on global regular functions.
If were an affine scheme, affine scheme reconstruction from global sections would make an isomorphism of schemes, because its global-section map is an isomorphism. But its image omits the closed point . 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.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 113 1 i Solution Created 2026-10-03 Updated 2026-10-05
On the principal open subscheme , the induced morphism of sheaves is the ring homomorphismIf is injective, this map is injective by exactness of localization: it is the localization of the injective -module map . Explicitly, if the displayed fraction vanishes, some power of annihilates ; injectivity gives , so the original fraction also vanishes. The principal opens form an open basis, so the kernel sheaf is zero on a basis and hence zero everywhere. Conversely, an injective morphism of sheaves induces an injection on global sections. On the whole affine scheme these sections are , exactly . Thus
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 113 3 ii Solution Created 2026-10-03 Updated 2026-10-05
For each point , choose a global section whose germ generates the rank-one free -module . The section is global, and tensor products commute with taking stalks. Thereforegenerates . Thus the tensor square is globally generated. More generally, the tensor product of globally generated line bundles is globally generated: choose a generator section for each factor at the point and tensor them. This works on every scheme, including nonreduced schemes.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 134 1 iii a Solution Created 2026-10-03 Updated 2026-10-05
Assume is ample. For every positive-dimensional integral subvariety , its restriction is an ample Cartier divisor. The asymptotic Riemann–Roch polynomial has leading termBy the Nakai–Moishezon criterion, the coefficient is positive, so this proves implication (a)(b).
For (a)(c), choose such that is very ample. Fix a closed point . A hyperplane through its image which does not contain the whole embedded gives a nonzero global section vanishing at . Such a hyperplane exists since . Its vanishing is therefore nonempty but not all of . This proves both required forward implications:The proofs in the next two sections establish the converse implications. Reduction to integral components is legitimate by ampleness on reduced components; in dimension zero every line bundle on a projective scheme is ample and there is no positive-dimensional subvariety to test.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 134 1 ii Solution Created 2026-10-03 Updated 2026-10-05
Put and let be the canonical global section cutting out the effective Cartier divisor . The divisor restriction exact sequence givesThe hypothesis says is an ample line bundle. By Serre vanishing, for all sufficiently large , so the long exact sequence in sheaf cohomology makessurjective for all such . These are finite-dimensional vector spaces. Their dimensions form a nonincreasing sequence of nonnegative integers, so the maps are isomorphisms from some point on. Exactness then implies that the restrictionis surjective for all sufficiently large .
Choose such an for which is also a globally generated line bundle. Lift a generating collection of its global sections to . At each point of , one lift has nonzero image in the one-dimensional residue-field fibre, hence generates the stalk of by Nakayama lemma. Outside , is nowhere zero and generates . Together these sections generate it everywhere. ThereforeThis proof works on an arbitrary projective scheme because the defining section of an effective Cartier divisor is a non-zero-divisor. If is empty, and the conclusion is immediate. Semiampleness is the conclusion: the pullback of a line avoiding the centre of a blowup of a smooth algebraic surface of satisfies the hypothesis on its support but has zero intersection with the exceptional curve, and therefore is not ample.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 134 2 iii Solution Created 2026-10-03 Updated 2026-10-05
Because and , its normal line bundle is . Thus every is trivial and has . The divisor restriction exact sequence gives, for every ,The maps are surjective. Their nonnegative finite dimensions eventually stabilize, so for every sufficiently large restriction is surjective. A lift of is nowhere zero along , and the canonical section of is nowhere zero outside . Together they generate . Therefore is semiample.
For those same large , the exact sequence of global sections givesThus , and the Iitaka dimension is exactly one:Equivalently, its basepoint-free multiple defines a morphism to a curve: it is nonconstant because its sections grow, and cannot have two-dimensional image because . This is the semiampleness of a square-zero rational curve; it uses no characteristic-zero vanishing theorem.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 118 2 d Solution Created 2026-10-03 Updated 2026-10-05
Evaluation is surjective as a sheaf morphism: near , any prescribed value is supplied by a constant holomorphic function. We therefore have a short exact sequence of sheaveswhere is the point ideal sheaf and the sheaf of holomorphic functions. By part (c), the skyscraper sheaf is flasque, so and for .
On the connected compact elliptic curve, the maximum modulus principle makes every global holomorphic function constant. Consequently , and evaluation on global sections is an isomorphism. Also : by Serre duality for compact complex manifolds, its dual is , generated by the nowhere-zero form descended from . Higher cohomology of vanishes by the Dolbeault theorem in complex dimension one.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 139 1 v Solution Created 2026-10-03 Updated 2026-10-05
First suppose is integral. Induct on its dimension of a scheme. The previous two parts handle all and also handle when eventually vanishes. Otherwise choose with a nonzero global section of . On an integral variety it defines an effective Cartier divisor , possibly empty, and multiplication by the section givesThus . For , the induction hypothesis bounds the last term by ; summing on each residue class modulo gives . For , is zero-dimensional and its positive-degree sheaf cohomology vanishes, so the same recurrence is bounded.
For a general projective scheme, a nonzero section can be a zero divisor, so that argument requires an additional step. The general cohomology growth for nef twists supplies it: for every coherent sheaf with support dimension ,Its proof uses Fujita vanishing, an ample section avoiding the associated points of , and induction on support dimension. Taking gives the required estimate for every , including nonreduced and reducible schemes; degrees above vanish.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 139 2 i b Solution Created 2026-10-03 Updated 2026-10-05
The relevant invariant of a divisor is its Iitaka dimension, also called its Kodaira dimension. If two linearly independent global sections occur in , their ratio is a nonconstant rational function. The sections are linearly independent, because a nonconstant rational function over an algebraically closed field is transcendental over that field. Thus some complete linear system of a divisor has positive-dimensional image, contradicting .
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 139 3 iii Solution Created 2026-10-03 Updated 2026-10-05
Fix any ample divisor . Choose so that both and have nonzero global sections, and choose . Condition (1) and the section subtraction lemma for big divisors yield an integer with . Adding an effective member of gives . Thus (1) implies the stronger condition (2').
If with ample and effective, multiplication by the section of injects into . The positive leading coefficient of the ample Hilbert polynomial supplies condition (1) along the infinite sequence . Condition (2) implies (3). Conversely, if , put . The Cartier divisor is numerically equivalent to , hence ample by Kleiman's criterion; the actual equality gives (2).
We have proved (1)(2')(2)(1), and (3)(2)(3). Also (2') implies (3'), while (3') implies (3). Hence all five conditions are equivalent. They characterize a big divisor; the ample-plus-effective formulation is Kodaira's lemma. The inconsistent use of and in the printed multiplier is resolved by using throughout.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 139 3 i Solution Created 2026-10-03 Updated 2026-10-05
Choose a very ample divisor such that has a nonzero global section; this is possible by taking a sufficiently high ample twist. Since is integral, multiplication by its th power injects into . By Serre vanishing and the Hilbert polynomial, . ConsequentlyThis is the polynomial bound for sections of a fixed divisor. The same proof works for a coherent sheaf of support dimension , by choosing the multiplying section to avoid its associated points; the bound is then .
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 118 2 c Solution 2026-10-03
Let be the sheaf of holomorphic -forms and the sheaf of smooth -forms. The Dolbeault-Poincaré lemma makesan exact resolution by fine sheaves, which are acyclic for global sections. Its global cochain complex computes Dolbeault cohomology, proving the Dolbeault theorem
Picard scheme of a curve 2026-10-14
For a smooth projective algebraic curve of genus , this smooth proper -dimensional scheme represents degree- line bundles in families. Charts come from nonspecial degree- bundles: their first sheaf cohomology vanishes, so the Riemann-Roch theorem gives exactly one independent global section, whose zero Cartier divisor varies algebraically. Twisting by fixed line bundles gives charts covering every degree- class. For degree zero the tensor product gives the Jacobian variety; other components are its torsors.
If is an effective Cartier divisor on a projective scheme and is ample, then is semiample. The divisor restriction exact sequence and Serre vanishing make surjective for large . Their finite dimensions stabilize, so restriction on global sections is eventually surjective. Lift generators on ; off , the canonical section of generates. Together these generate , including on nonreduced .
For a degree- divisor class with , let be its unique effective representative. Under , evaluation of the unique global section gives , including multiplicities. If , every evaluation kernel is nonzero and the entire curve lies in the theta divisor. These restrictions also show that : after pulling back to a dominant product of curves, the ratio of any section to the defining section of the theta divisor is constant on each general coordinate curve, hence constant everywhere.
Riemann singularity theorem 2026-10-14
For a compact Riemann surface of genus , the theta divisor in has the displayed multiplicity. A local determinant description of the theta divisor reduces to an by matrix vanishing at , where . Its linear term is cup product with a tangent class in . The invertible cup-product direction for a line bundle on a curve shows that its determinant has order exactly . Consequently the singular locus consists precisely of classes with at least two independent global sections.