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 125 2 i Solution Created 2026-10-03 Updated 2026-10-06
For an elliptic curve over , where is a prime power, the Hasse theorem for elliptic curves isWe prove it from the degree form on elliptic-curve endomorphisms, including the required degree facts.
A nonzero homomorphism of elliptic curves is a finite surjective morphism; its degree is the degree of the induced extension of function fields. Set the degree of the zero homomorphism to . The degree of an isogeny is also the degree of the pullback of any point divisor: locally, a finite map of smooth curves gives a module of rank equal to the function-field degree, and its fibre length is the sum of the local multiplicities. For a composition, the tower law for function fields givesIf the isogeny of elliptic curves is separable, its pullback of a nonzero invariant differential on an elliptic curve is nonzero. Translation invariance implies that its differential is nonzero at every point, so each fibre point has multiplicity . Every fibre is a translate of the kernel. ThereforeFor an inseparable isogeny, multiplicities must be retained; counting geometric kernel points alone would be incorrect.
Here is a divisor proof of the degree parallelogram law. On , let be the diagonal and the locus . A Weierstrass -function has a double pole at , and the equation describes or . ConsequentlyThis remains valid in characteristic using a general Weierstrass equation of an elliptic curve: the -map still has degree and its two points are exchanged by negation. The divisor identity gives an identity of line bundles. Pull it back by and take degrees. The diagonal is the inverse image of under subtraction and the antidiagonal under addition, soUsing line bundles makes this pullback argument valid even when , or one map is zero, when directly substituting into the rational function would fail.
Let . Applying the degree parallelogram law to and gives , with . Induction givesMore generally, the same recurrence and polarization show that the degree restricted to the subgroup generated by two endomorphisms is a quadratic form: its mixed coefficient is determined by their sum or difference. This can be checked directly by the second-difference recurrence in each integer variable and the parallelogram identity in the two diagonal directions.
Let be the -power Frobenius isogeny of an elliptic curve. It has degree : it is purely inseparable, has one geometric point in each fibre, and sends a local parameter at the rational point to its th power, giving fibre multiplicity . Its pullback of any differential is zero. On the other hand, the standard addition rule for an invariant differential on an elliptic curve givesHence is separable, and its kernel consists exactly of the points fixed by Frobenius, namely . Put and . Thenand the quadratic-form calculation gives, for all ,If , the real polynomial is negative on an open interval. That interval contains a rational with , contradicting the displayed nonnegativity after multiplication by . Thus , which is precisely the Hasse theorem for elliptic curves.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 125 4 Solution Created 2026-10-03 Updated 2026-10-05
Galois cohomology and the finite division quotient. Let be a number field, with its profinite topology, and a discrete continuous -module. Its Galois cohomology is continuous group cohomology. In degrees zero and one,Thus a crossed homomorphism measures the failure of a choice to be Galois-invariant, and changing the choice changes the cocycle by a coboundary. A short exact sequence of such modules gives a long exact sequence in group cohomology. The connecting map sends an invariant element to the cocycle obtained by lifting it and comparing its Galois translates.
For an elliptic curve and , multiplication by is surjective on and has kernel . The Kummer exact sequence of an elliptic curve therefore givesExplicitly, choose and set . Replacing changes this by a coboundary; replacing by , , does not change it. If its class is zero, some torsion translate of is Galois-invariant, so . This proves injectivity of the Kummer map of an elliptic curve.
The whole group need not be finite. The crucial restriction is ramification. Let contain the finite primes of bad reduction of an elliptic curve and those dividing . At , good reduction and prime-to-residue-characteristic multiplication imply that every division point of a local point is in an unramified extension, by the formal-group lifting argument. Thus every global Kummer class restricts trivially to inertia there.
Choose a finite Galois extension containing and the th roots of unity, and enlarge to include its ramified primes. Over , the torsion module is constant and can be identified with . By Kummer theory,For classes unramified outside the primes above , valuations outside must be divisible by . Their restriction images therefore lie in , the square of an S-unramified power class group. Its finiteness can be seen without assuming the Mordell-Weil theorem: the finiteness of S-unramified Kummer classes follows fromThe S-unit group is finitely generated by the Dirichlet unit theorem and its valuations at . The localized ideal class group is a quotient of the ordinary ideal class group of , which is finite by the finiteness of the ideal class group. The Minkowski bound for ideal classes supplies an integral ideal of bounded norm in each class, and only finitely many integral ideals have bounded norm.
The kernel of restriction from to is also finite: a cocycle trivial after restriction can, after subtracting a coboundary, be made to factor through the finite group . There are only finitely many maps from that group to the finite module . This is the finite-extension kernel of a Kummer map mechanism. Consequently the possible Kummer classes form a finite set, provingThis is the Weak Mordell-Weil theorem. Imposing membership in the local Kummer image at every place refines the finite unramified collection to the n-Selmer group. Its relation with the obstruction to a globally rational point iswhere the last group is the -torsion of the Tate–Shafarevich group. Thus local solubility gives a finite computable upper bound but can leave a genuine global obstruction.
Heights and finite generation. For the other essay, normalize the absolute values on a field so that the product formula holds. The Absolute logarithmic Weil height of isAt the infinite places the local degrees are one or two and the usual real or complex modulus is used. The product formula makes this independent of scaling; it is also unchanged by extending . Over , using coprime integral coordinates gives , the naive height on the projective line. The Northcott theorem says that points of bounded height and bounded field degree form a finite set; in particular this holds over the fixed field . One way to see the finiteness is the height-Mahler measure formula: for an algebraic number of degree , the coefficients of its primitive minimal integer polynomial are bounded by binomial factors times . Bounds on and therefore leave only finitely many integer polynomials and hence only finitely many algebraic numbers.
Set for , and . Since has degree two, bounded gives only finitely many points of . The height growth under a morphism of the projective line states that a degree- morphism gives . The upper bound follows from its homogeneous coordinate polynomials; the lower bound follows from their having no common zero, using a resultant identity at each place. Applied to the degree-four duplication map, it gives a uniform constant withConsequently is Cauchy: the absolute difference of consecutive terms is at most . Define the canonical height of an elliptic curve byIt is nonnegative and satisfies . The factor one-half is the conventional normalization for the divisor , since the first-coordinate height belongs to .
The addition-divisor identity also gives the bounded-error height identityuniformly in . This is the height form of the identity for the sum and difference pullbacks of the line bundle associated to ; it includes points where the affine addition formulas have a zero denominator. Apply it to , divide by , and take the limit. Thus satisfies the parallelogram law, and its polarization is a bilinear form, the height pairing. Nonnegativity implies the Cauchy-Schwarz inequality for that pairing, by evaluating its nonnegative quadratic polynomial on for all integers and approximating real ratios by rational numbers. In particularThese properties do not presuppose finite generation. Bounded canonical height gives a finite set by the bounded difference and the Northcott theorem. Torsion points have canonical height zero. Conversely, if , all have bounded naive height, so two coincide; their difference makes torsion. Thus the height detects precisely the free part, once finite generation is proved.
Use the Weak Mordell-Weil theorem with and choose finitely many coset representatives . Write , and put . The height inequality yieldsIteration enters the finite set , since after steps the height is at most . Unwinding expresses every point using that finite set and the . This proves the canonical-height proof of Mordell-Weil finite generation and henceThe Fundamental theorem of finitely generated abelian groups supplies this decomposition. The Mordell-Weil theorem combines a finite quotient from arithmetic Galois cohomology with a contracting height descent; either ingredient alone would not establish finite generation.
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 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 2017 iii Paper 134 3 iii b Solution Created 2026-10-03 Updated 2026-10-05
Choose an ample Cartier divisor . Put initially , for small positive real . Since is nef, the nef-plus-ample ampleness lemma makes ample. The polynomialhas . Thus the desired strict inequality holds when are sufficiently small and positive.
We must also arrange rationality of the two specified classes; itself need not be rational. Choose rational ample classes and sufficiently near and , and defineThen is close to and is close to , so both are ample real divisors by openness of the ample cone. Also and are rational and ample. Continuity preserves the strict inequality, giving
Here is the needed algebraic Morse inequality for ample divisors, with its section-count proof. Choose rational Cartier divisor representatives of and a common positive integer making very ample integral Cartier divisors. For the section-count argument rename these scaled representatives ; undoing this scaling restricts section indices to sufficiently divisible multiples and leaves bigness unchanged. Choose an effective Cartier divisor by taking a defining section that avoids the associated points of . Repeated divisor restriction exact sequences giveBecause is very ample, for each a section of avoiding the finitely many associated points of gives an injection into . Thus every summand is at most . By Serre vanishing and asymptotic Riemann–Roch for the ample ,For , the restriction term is the constant length of , giving the same formula directly. The positive coefficient proves that , hence , is big. Scaling back preserves bigness, so
For a general projective scheme, enforce the same strict inequality separately on each positive-dimensional reduced irreducible component , using its own dimension . At every such expression equals the positive number . Finitely many conditions are preserved by one sufficiently small choice and one sufficiently close rational approximation on . The top-dimensional inequalities imply the printed inequality for with its positive generic multiplicities; the section proof on each component makes componentwise big. This avoids inferring bigness on every component from just a positive sum. The displayed inequality is used for . For , its literal intersection power is undefined; handle this vacuous positivity case separately. Every line bundle is ample, all numerical classes are zero, and the componentwise bigness convention makes the conclusions automatic.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 134 3 i Solution Created 2026-10-03 Updated 2026-10-05
An ample real divisor is a finite positive real combination of ample Cartier divisors:Equivalently its numerical class lies in the ample cone. This is a numerical condition even when the coefficients are irrational; it does not mean that some integer multiple of must be an integral divisor.
On an integral projective variety, a big real divisor is a finite positive real combination of big Cartier divisors. Equivalently, by the real form of Kodaira's lemma,For an integral Cartier divisor, bigness means maximal section-growth order along sufficiently divisible positive , or Iitaka dimension . The real linear equivalence of divisors formulation permits finite positive combinations of effective Cartier divisors, whose supports are codimension one. The definitions and the ample-plus-effective formulation on integral varieties are discussed in Fujino's notes on big real divisors.
For the paper's assertions on a general projective scheme, use componentwise bigness on a projective scheme: require bigness on every reduced irreducible component. All arguments below can then be carried out on those finitely many integral components; ampleness is also detected there. On a reducible scheme, merely asking for maximal total section growth on one component is insufficient. For example , with restricting to on the first component and on the second, has quadratic total section growth, but negative intersection with every line in the second component. No finite collection of codimension-one subvarieties can contain all those lines. Thus that weaker meaning would make part (ii) false. In dimension zero the positivity statements are vacuous and every line bundle is ample; the compatible bigness convention also regards it as big.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 113 3 a Solution Created 2026-10-03 Updated 2026-10-05
First consider the preliminary cohomological facts. The exam's term flabby means flasque sheaf: every restriction map is surjective. For open sets , lift a section of to using the given surjectivity when is flasque. Extend that lift to using the flasque sheaf , then project to . This proves that is a flasque sheaf.
One construction of a flasque resolution starts with the sheaf , with pointwise restriction. It is a sheaf of abelian groups with surjective restrictions, and sending a section to all its germs embeds in it. Repeating on the cokernels constructs an exact sequencewith each flasque. Define sheaf cohomology by the cohomology of the cochain complex of global sections of such a resolution; the groups are independent of the chosen flasque resolution. If is itself flasque, the quotient result just proved makes every successive cokernel flasque. Applying the assumed surjectivity of sections to each successive short exact sequence of sheaves proves exactness of the global-section complex in positive degrees. Thus for .
Now let be the function field of the smooth algebraic curve. Its divisor class group isHere is the discrete valuation of the discrete valuation ring , and consists of the principal divisor elements . These sums have finite support: on a finite affine cover, represent as a fraction of regular functions; each nonzero numerator or denominator has only finitely many zeros on a curve, since a proper closed subset of a Noetherian one-dimensional irreducible space is finite.
For a divisor on an algebraic curve , define the line bundle associated to a divisor byIn this expression the valuation condition applies to nonzero . If is a uniformizer, the stalk is . Shrinking around removes all other zeros and poles of and all other points in the support of , so this also gives an actual local generator. Thus is a locally free sheaf of rank one. Multiplication gives an isomorphismon every stalk. If , then , so the construction factors through a homomorphism .
It is injective: if is trivial, an isomorphism from supplies a nonzero global rational generator . At each closed point, generates , giving , hence .
It is surjective: take a line bundle and a nonzero rational section , obtained by choosing a frame on any nonempty trivializing open set. If is a local frame, write with . Since changes of frame are units, is independent of the frame near . A finite trivializing cover shows that these numbers have finite support. Set . The map gives , because at the condition is exactly regularity of . Replacing by changes by , so this construction is well defined on classes. This proves the divisor class group and Picard group of a smooth curve identification
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 113 3 c Solution Created 2026-10-03 Updated 2026-10-05
The sheaf of nonzero rational functions on an irreducible variety is a flasque sheaf. Indeed, its sections on any nonempty open subset are : irreducibility makes nonempty open subsets connected and makes any two of them meet, so locally constant rational-function values must agree. Restrictions between such opens are identities; restriction to the empty set is also surjective. The preliminary argument therefore givesApply the long exact sequence in sheaf cohomology to the short exact sequence of sheaves of multiplicative abelian groupsIts relevant part isPart (b) identifies the first map with the principal divisor map . Exactness therefore identifies its cokernel with . Combining this with part (a) givesThe connecting map sends a divisor to the class of the unit ratios of its local rational representatives; these are precisely the transition data of the associated line bundle, up to the usual choice of inverse transition convention.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 113 4 Solution Created 2026-10-03 Updated 2026-10-05
The resolution principle for sheaf cohomology states that an exact sequencewith for every computes as the cohomology of . In particular one may use a flasque resolution.
An affine morphism is one for which the inverse image of every affine open subset of is affine. We use vanishing of quasi-coherent cohomology on an affine scheme: a quasi-coherent sheaf has zero higher sheaf cohomology on an affine variety. Take a flasque resolution . Each direct image sheaf is flasque, because restriction on is restriction on . On an affine open , the augmented complex of sections isIts positive-degree cohomology vanishes by the affine theorem, and its degree-zero kernel is . Checking exactness on this affine basis shows that is a flasque resolution of . The global-section complexes are identical, proving cohomology under an affine morphism:
For projective space, put with its usual grading and . Construct the twisting sheaf on projective space by setting its sections on to be the degree- part , viewed as a module over . It is free of rank one with generator , even when is negative. On overlaps the generators differ by the unit , so these modules glue to a line bundle .
Assume first . These local modules all embed in the graded fraction field, and agreement on overlaps identifies global sections with . Since is a unique factorization domain and at least two distinct variables occur, a reduced fraction belonging to every has no nonconstant denominator. Thus the intersection is . Counting monomials by stars and bars givesHere . The printed formula needs the convention that a counting binomial coefficient is zero when its upper integer is smaller than its nonnegative lower integer; the generalized polynomial convention for negative upper integers would give incorrect dimensions. For , every twist on the single point is trivial, so for every and its higher cohomology is zero. This case supplies the base of dimension induction; the displayed top-cohomology formula below is for .
The standard affine opens have affine finite intersections. The acyclic cover theorem and affine vanishing identify sheaf cohomology with a Čech cochain complex ending in degree . In particular for .
Let , with inclusion . Multiplication by gives the hyperplane exact sequence for twisting sheavesExactness is checked in the local trivializations, where a hyperplane equation is a non-zero-divisor and the quotient is its restriction to . Since a closed immersion is an affine morphism, the first part identifies the cohomology of with that on .
For , the long exact sequence in sheaf cohomology and giveStarting from the given vanishing for sufficiently large and descending gives for and for .
Now let , and assume the formulas in dimension . Restrictionis surjective for every : for it is polynomial restriction, and for both groups are zero since . The long exact sequence in sheaf cohomology therefore makes injective. For , dimension induction gives , so the same sequence makesinjective. Composing these injections up to a sufficiently large twist, where the target is zero by the given hypothesis, proves for every and .
The remaining part of the long exact sequence in sheaf cohomology is consequentlywhere the last zero uses the affine-cover dimension bound on . Hence . Using dimension induction and eventual vanishing, downward summation yieldsby telescoping Pascal's identity; for it is zero. Thus, without invoking Serre duality in the induction,Finally, the canonical bundle of projective space is , obtained by taking determinants in the dual Euler sequence. Serre duality therefore predictsIt pairs the two computed extreme-degree dimensions and pairs zero intermediate groups with zero intermediate groups, exactly as required.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 139 2 ii Solution Created 2026-10-03 Updated 2026-10-05
Because has no fixed components, choose two members with no common component. This uses the infinitude of the algebraically closed field and avoidance of finitely many proper linear subspaces of the section space. On a smooth algebraic surface, their intersection number is the sum of local intersection multiplicities. Since , they are disjoint. A basepoint of would belong to both, so is basepoint-free.
The resulting morphism has nonconstant image since . Its image cannot be a surface: a generically finite morphism defined by would make the positive product of its degree and the degree of its image. Thus the image is a curve. Apply Stein factorization to obtain with connected fibers and , where is a smooth projective curve. There is a positive-degree line bundle on with . The projection formula for sheaves and Riemann-Roch theorem give . Therefore
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 159 1 Solution 2026-09-28
The Picard group is the set of isomorphism classes of line bundles on , with tensor product as addition, the trivial bundle as zero and the dual bundle as inverse. On a smooth projective surface, line bundles can be represented by divisors. The intersection pairing on the Picard group of a surface is the symmetric bilinear mapobtained by moving the divisors into proper position and counting their intersections with multiplicity.
The surfaceis the first Hirzebruch surface . If is its negative section and a fiber of the ruling, its Picard lattice of a Hirzebruch surface is
Now let be the given nonisomorphic birational morphism. A birational morphism between smooth projective surfaces factors as a nonempty sequence of point blowups. Let , and take the total transform on of the exceptional curve of the first blowup. The intersection formula for blowing up a surface givesTherefore the nonzero Picard group element satisfiesThis is the isotropic divisor from a nontrivial birational morphism to the projective plane.
For a morphism , the pullback of the hyperplane bundle has the formfor an integer . If a line is contracted to a point, this bundle restricts trivially to , whereasIts degree is therefore zero, so . The homogeneous sections defining are then constants, and is constant. This proves the morphism from the projective plane contracting a line criterion.
Finally choose an integer and blow up distinct points of the smooth curve . For the resulting morphism , let be the exceptional curves. The strict transform isand the self-intersection after blowing up points on a smooth curve formula givesBlowing up a smooth point of a smooth curve does not change that curve itself, so is an isomorphism.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 113 2 d Solution 2026-09-28
The standard affine charts of are copies of , and the product charts of are copies of . Their local rings are localizations of polynomial rings over and hence are regular local rings. Both schemes are therefore regular.
On a regular integral scheme every Weil divisor is Cartier, so the divisor class group is naturally the Picard group. Pullback of line bundles along the Segre embedding therefore definesThe two groups areThe Segre coordinates are bihomogeneous of bidegree , so . In these bases the map is and
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 113 1 b Solution Created 2026-09-24 Updated 2026-09-25
On the standard charts and of , a section of the twisting sheaf on projective space is represented by a degree-zero element of the corresponding localization of . A global section is therefore a homogeneous polynomial of degree when . There are no nonzero global sections for . Hence
Local isomorphism on every member of a fixed affine cover does not imply a global isomorphism: the local identifications may have different transition functions. For example, and are both trivial on the two standard affine charts of , but they are not isomorphic because their spaces of global sections have dimensions one and two.
An injective map between line bundles need not be an isomorphism. Multiplication by a nonzero section giveson ; its cokernel is a nonzero skyscraper sheaf supported at the zero of the section.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 126 4 i Solution Created 2026-09-24 Updated 2026-09-25
The Theorem of the square says that for an abelian variety , a line bundle , and ,where denotes translation by .
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 152 1 c Solution Created 2026-09-24 Updated 2026-09-25
On , let be the negative section of a Hirzebruch surface, so , and let be the fiber class of a Hirzebruch surface, with and . The line bundle is very ample. Indeed, the toric ampleness criterion for isand on a smooth complete toric variety every ample line bundle is very ample. These inequalities hold for .
Explicitly, after choosing the standard lattice coordinates for , its lattice polytope of a toric divisor isThe monomials indexed by the lattice points of this polygon separate torus orbits and tangent directions, so their Kodaira map is a closed embedding. This directly verifies that the resolved proper toric surface is projective.
Picard charts from nonspecial divisors 2026-10-14
A degree- line bundle with vanishing first sheaf cohomology has exactly one section up to scale, by Riemann-Roch theorem. Its zero divisor is unique and varies in families by cohomology and base change for line bundles on a curve. For a fixed divisor of degree , the open subset of where has vanishing first cohomology gives a chart in degree , by . Different charts are glued by the unique residual divisor of wherever that line bundle is again nonspecial. Normalizing the universal bundles along a fixed rational point gives their canonical compatible gluing and a representing scheme for the Picard functor of a curve.
Picard functor of a curve 2026-10-14
For a smooth projective geometrically connected algebraic curve , this functor assigns to a scheme the line bundles on having degree on each geometric fibre, modulo bundles pulled back from . One generally takes the associated sheaf for faithfully flat covers. A rational point of permits normalization of every line bundle along that point; normalized bundles have no automorphisms preserving their normalization and satisfy effective descent. The resulting functor is represented by the degree- component of the Picard scheme of a curve.
Picard group Created 2026-09-24 Updated 2026-09-24