Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 80 (Cohomology of Coherent Sheaves)

At a point $P$ of the smooth algebraic curve $X$, the local ring ${\mathcal{O}}_{X,P}$ is a discrete valuation ring. Embed the projective variety $Y$ in ${\mathbb{P}}^N$. A rational map of algebraic varieties to $Y$ has homogeneous coordinates $[f_0:\cdots :f_N]$ in the function field $k(X)$, not all zero. Let $v_P$ be the local valuation, choose a uniformizer $u$, and set $m={\min }_{f_i\ne 0}{v}_P(f_i)$. The coordinates $u^{-m}{f}_i$ are regular at $P$, and at least one is a unit. On a sufficiently small open subset containing $P$, all these functions are regular and that coordinate stays nonzero. Their ratios therefore define a morphism of algebraic varieties into the corresponding affine chart of ${\mathbb{P}}^N$. Every homogeneous equation defining $Y$ vanishes on the original dense open, hence vanishes identically on this neighborhood. The local extension consequently lands in $Y$.

Two such extensions agree wherever the original rational map of algebraic varieties is defined. Their equalizer is closed, because $Y$ is a separated variety, and contains a dense open. Thus they agree throughout their overlap and glue. The rational map has a unique extension to a morphism on all of $X$. Projectivity is needed for the target; the source need not be projective.

For the function field correspondence, let $K/k$ be a finitely generated field extension of transcendence degree one. Choose a transcendence basis consisting of one element $t$. The extension $K/k(t)$ is finite: finitely many algebraic generators each have finite degree. Take the integral closures of $k[t]$ and $k[t^{-1}]$ in $K$. The finiteness assumption allowed in the paper makes both finite modules over their original coordinate rings. Integral closure commutes with localization, so these two rings agree over $k[t,t^{-1}]$ and glue to the finite normalization model of a one-variable function field $C_K\to {\mathbb{P}}^1$. This is a finite morphism, hence projective, and its function field is $K$. Its one-dimensional normal local rings are discrete valuation rings, so it is regular. Since the algebraically closed field $k$ is perfect, regularity here implies smoothness. Thus $C_K$ is a smooth projective curve.

Conversely the function field of a smooth projective curve is finitely generated over $k$ and has transcendence degree one. Two such curves with isomorphic function fields have mutually inverse birational maps. The extension argument just proved makes both maps morphisms; their composites equal the identity on a dense open, and hence everywhere by separatedness. This proves uniqueness of the smooth projective model. More precisely, pullback gives a contravariant correspondence between nonconstant morphisms of smooth projective curves and $k$-embeddings of their function fields. A field embedding defines the map on affine coordinates near the generic point, and the extension argument completes it. Constant morphisms are excluded from this correspondence because they do not induce embeddings of function fields.

Finally let $f:C\to D$ be nonconstant. Its image is dense, and pullback gives a finite extension $k(D)\subset k(C)$: both fields have transcendence degree one and the latter is finitely generated. Normalize $D$ in $k(C)$, applying the allowed finite-integral closure result on affine charts. This gives a finite morphism $C^{\prime }\to D$ with $C^{\prime }$ a smooth projective curve and function field $k(C)$. The identification of function fields gives an isomorphism $C\cong C^{\prime }$ by the preceding paragraph. Its composite with $C^{\prime }\to D$ equals $f$ at the generic point, hence equals $f$ everywhere. Every nonconstant morphism of these curves is finite.

Write the affine variety as $X=\mathrm{Spec}A$. Its coordinate ring is Noetherian, and a coherent sheaf is $\mathcal{F}=\tilde{M}$ for a finitely generated $A$-module $M$. We first justify the acyclic resolution used to compute its sheaf cohomology.

If $I$ is an injective module, then $\tilde{I}$ is a flasque sheaf. Indeed every open $U$ in this Noetherian scheme has the form $X\setminus V(\mathfrak{a})$ for a finitely generated ideal. Clearing denominators givesHere a homomorphism $h:{\mathfrak{a}}^n\to I$ gives, on a principal open $D(f)$ with $f\in \mathfrak{a}$, the fraction $h(f^n)/f^n$; the fractions agree on overlaps. Conversely choose fractions for a section on a finite principal-open cover. A common power clears their denominators and the finitely many compatibility relations. Multiplication by a further power of $\mathfrak{a}$ kills any elements supported on $V(\mathfrak{a})$, producing a well-defined map on some ${\mathfrak{a}}^n$. Changing that power accounts for the direct limit. This is the sheafification of an injective module is flasque construction. Since $I$ is injective, each map ${\mathfrak{a}}^n\to I$ extends to $A\to I$. Thus every section on every open extends to $X$. Extending globally and restricting again also proves surjectivity for restrictions between arbitrary opens.

Choose an injective module resolution $0\to M\to I^0\to I^1\to \cdots$. Localization is exact, so sheafifying gives a flasque resolution of $\mathcal{F}$. Its global-section complex is exactly $I^{\bullet }$, since $\Gamma (\mathrm{Spec}A,\tilde{I})=I$. This complex is exact in positive degrees, and a flasque resolution computes sheaf cohomology. ConsequentlyThe argument actually proves vanishing of quasi-coherent cohomology on an affine scheme, without finite generation of $M$.

For the intersection assertion, varieties are taken to be separated, as usual. The diagonal morphism $X\to X{\times }_{k}X$ is a closed immersion. Its inverse image over $U{\times }_{k}V$ identifies $U\cap V$ with a closed subvariety of the affine variety $U{\times }_{k}V$. A closed subvariety of an affine variety is affine: its coordinate ring is the ambient coordinate ring modulo its defining ideal. Thus $U\cap V$ is affine. Repeated application gives affine finite intersections. Without separatedness this statement need not hold.

Choose a finite affine open cover $(U_{\alpha })$ of the projective variety $X$. For ordered indices write $U_{{\alpha }_0\cdots {\alpha }_p}=U_{{\alpha }_0}\cap \cdots \cap U_{{\alpha }_p}$. The answer is the cohomology of the Čech cochain complexIn particular $H^0=\ker (C^0\to C^1)$ and, in positive degree,The chart modules $\Gamma (U_{\alpha },\mathcal{F})$, together with their restriction and gluing maps, supply the modules on overlaps by localization. The gluing maps are essential; abstract chart modules alone do not specify a sheaf.

For completeness, take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$ and its double complex of Čech cochain complexes on this cover. In the resolution direction, its cohomology is $H^q(U_{{\alpha }_0\cdots {\alpha }_p},\mathcal{F})$, which vanishes for $q>0$ by the affine result. In the cover direction, the augmented Čech cochain complex of a flasque sheaf is exact: sections can be extended from overlaps, successively adjusting a cocycle to eliminate one cover member at a time. Equivalently the augmented sheaf complex is stalkwise contracted by inserting a cover index containing the point. Thus computing the total cohomology in the two directions identifies Čech cohomology with sheaf cohomology, proving the displayed description.

Choose a nonconstant element $t$ of the function field $k(C)$. By Question 1 it gives a finite morphism $\pi :C\to {\mathbb{P}}^1$. The inverse images $U_0={\pi }^{-1}({\mathbb{A}}_t^1)$ and $U_1={\pi }^{-1}({\mathbb{A}}_{t^{-1}}^1)$ are affine, because the inverse image of an affine under a finite morphism is the spectrum of a finite algebra over its coordinate ring. They cover $C$. This proves the two-affine cover of a smooth projective curve assertion.

Their intersection is affine by Question 2. Thus the Čech cochain complex for any coherent sheaf $\mathcal{F}$ has just two nonzero terms,It follows that

On the projective line, put $t=X_1/X_0$ on the first chart. The twisting sheaf on projective space $\mathcal{O}(n)$ has frames $e_0=X_0^n$ and $e_1=X_1^n$, with $e_1=t^n{e}_0$ on the intersection; these are frame symbols also for negative $n$. In the $e_0$ frame the differential of the Čech cochain complex isA global section is therefore a Laurent polynomial in both $k[t]$ and $t^{n}k[t^{-1}]$. Its monomials have exponents $0\le j\le n$. For $n\ge 0$ the sections have basis $e_0,t{e}_0,\dots ,t^n{e}_0$; for $n<0$ there are none. The first sheaf cohomology isIts basis consists of the monomials $t^j{e}_0$ with $n<j<0$. These are precisely the exponents missed by the two chart spaces. Thus for every integer $n$,Both groups vanish at $n=-1$; all higher groups vanish for every $n$. This is also an explicit calculation of the Čech cohomology of twists on the projective line.

The printed vanishing claim needs the restriction $i\ge 1$. On $C={\mathbb{P}}^1$ with $\mathcal{F}=\mathcal{O}$ and $D=n\{\infty \}$, Question 3 gives $h^0(\mathcal{F}(D))=n+1$ for every $n\ge 0$. Thus no bound on the degree of a divisor can force vanishing in degree zero. We prove finiteness in every degree and the corrected vanishing in every positive degree, uniformly over all divisors on an algebraic curve of large degree.

First take a finite morphism $\pi :C\to {\mathbb{P}}^1$ as in Question 3 and put $E={\pi }_{\ast }\mathcal{F}$. The direct image sheaf $E$ is coherent: on an affine chart its module is finitely generated over a finite algebra and hence over the chart's coordinate ring. The two affine-cover Čech cochain complexes for $\mathcal{F}$ on $C$ and for $E$ on ${\mathbb{P}}^1$ are identical. Therefore

We need an elementary presentation of any coherent sheaf $E$ on the projective line. Choose finitely many generators on each of its two standard affine charts. A generator on the first chart restricts to an element of the second chart's module localized at $t^{-1}$. Multiplication by a sufficiently large power of $t^{-1}$ brings that element into the second chart's module. Since the transition of $\mathcal{O}(n)$ is $t^n$, this makes the original generator extend to a global section of $E(n)$. A generator on the second chart similarly extends after clearing its denominator at $t=0$. Use one common exponent for all the finitely many generators. The extended sections generate $E(n)$ on both charts. Untwisting gives a surjectionwith coherent kernel $K$. The long exact sequence in sheaf cohomology and Question 3 show that $H^1(E)$ is a quotient of the finite-dimensional $H^1(P)$, since $H^2(K)=0$. Apply the same presentation argument to $K$ to obtain finiteness of $H^1(K)$. Then the segment $H^0(P)\to H^0(E)\to H^1(K)$ proves finiteness of $H^0(E)$. Higher groups vanish by Question 3. Hence every $H^i(C,\mathcal{F})$ is finite-dimensional.

Next establish a useful degree estimate without using the duality requested later. A global regular function on $C$ is constant: a nonconstant one would give a finite morphism $C\to {\mathbb{P}}^1$ by Question 1. Such a morphism is surjective, because its closed image contains the generic point, but a regular function misses $\infty$, a contradiction. Thus $H^0(C,{\mathcal{O}}_C)=k$. Write $g=h^1(C,{\mathcal{O}}_C)$, the genus of a smooth projective curve. For any divisor on an algebraic curve $A$ and point $P$, the local discrete valuation ring givesThe last term is a length-one skyscraper sheaf. Additivity of the Euler characteristic of a coherent sheaf and repeated addition or subtraction of points giveThe inequality follows because $h^1$ is nonnegative, and is sufficient for the following argument.

Fix $P\in C$. For $n\ge g+1$ this estimate gives at least two independent sections of ${\mathcal{O}}_C(nP)$, one of them the constant section. A nonconstant section is a rational function on an algebraic variety whose poles are confined to $P$. It must have a pole there, since a globally regular function is constant. Its extension $C\to {\mathbb{P}}^1$ is finite, and its inverse image of ${\mathbb{A}}^1$ is exactly $C\setminus \{P\}$. Consequently $U=C\setminus \{P\}$ is affine.

Let $T$ be the torsion subsheaf of $\mathcal{F}$ and let $V=\mathcal{F}/T$. The support of $T$ is finite; such a coherent sheaf is a finite direct sum of sheaves supported at points and has zero higher sheaf cohomology. The torsion-free sheaf $V$ is locally free: a finitely generated torsion-free module over a discrete valuation ring is free. Tensoring by any invertible sheaf is exact, so

The inclusions $V(nP)\subset V((n+1)P)$ have point-supported quotients. Their long exact sequences in sheaf cohomology therefore give surjections on $H^1$. Moreovera section regular away from $P$ has some finite pole order at $P$. Compute this direct limit on a finite affine cover of $C$. All its intersections with $U$ are affine by Question 2, and localization commutes with filtered direct limits. The Čech cochain complex consequently givesThe maps from $H^1(C,V)$ to these groups are surjective. Choose a finite basis of this starting vector space. Each basis vector becomes zero at some stage in the direct limit; one common stage kills the entire basis. Thus there exists $n_0\ge 0$ with $H^1(C,V(n_{0}P))=0$.

Now let $D$ be any divisor on an algebraic curve with $\mathrm{deg}D\ge n_0+g$. The estimate above gives $h^0({\mathcal{O}}_C(D-n_{0}P))\ge 1$. A nonzero section identifies $D-n_{0}P$ with an effective divisor $E$ up to linear equivalence of divisors. Hence $V(D)\cong V(n_{0}P+E)$. Adding $E$ gives a point-supported quotient of $V(n_{0}P+E)/V(n_{0}P)$, so its first sheaf cohomology is a quotient of $H^1(C,V(n_{0}P))=0$. We have proved uniform high-degree vanishing on a smooth projective curve:This bound concerns all high-degree divisors, not only multiples of a previously chosen divisor.

We prove the required duality pairing explicitly. This also identifies the word “canonical” in the final conclusion: the isomorphism will be the intrinsic residue trace on a smooth projective curve, not a choice of basis for a one-dimensional vector space.

Begin on the projective line. With $s=t^{-1}$, the equation $ds=-t^{-2}dt$ identifies its sheaf of Kähler differentials over a field with $\mathcal{O}(-2)$. The Laurent description in Question 3 pairs $H^1(\mathcal{O}(a))$ with $H^0(\mathcal{O}(-a-2))$ by multiplication and extraction of the coefficient of $t^{-1}dt$. For $a\le -2$ the respective bases areand their pairing matrix has entriesThis is an invertible antidiagonal matrix. For $a\ge -1$ both spaces vanish. The coefficient extraction is well defined on the Čech cohomology quotient: a differential regular on the first chart has no $t^{-1}dt$ term, and one regular on the second chart is a sum of terms $t^{-2-j}dt$ with $j\ge 0$, also with no such term.

To extend this calculation to every vector bundle $E$ on the projective line, we give the splitting argument. Every line bundle on ${\mathbb{P}}^1$ is $\mathcal{O}(a)$: choose a nonzero rational section to obtain a divisor, and use the divisors of $t-c$ to move each finite point to infinity. The denominator-clearing argument in Question 4 makes both $E(n)$ and $E^{\vee }(n)$ globally generated for large $n$. The first assertion ensures $H^0(E(-a))\ne 0$ for some integer $a$; the second gives an embedding $E\hookrightarrow \mathcal{O}(n{)}^{\oplus r}$, so this is impossible for $a>n$. Choose the largest such integer $a$.

A nonzero section gives $\mathcal{O}(a)\to E$. Saturating its image gives a line subbundle; if it were strictly larger, it would be $\mathcal{O}(a+d)$ for a positive integer $d$, recording the section's zeros, contrary to maximality. Thus the quotient $F$ is a vector bundle. Induct on rank and write $F={⨁}_j\mathcal{O}(b_j)$. Twist the sequence by $-a-1$. Since $H^0(E(-a-1))=0$ and $H^1(\mathcal{O}(-1))=0$, exactness forces $H^0(F(-a-1))=0$, whence every $b_j\le a$. Local lifts of a quotient summand differ on overlaps by a Čech cocycle in $\mathcal{O}(a-b_j)$. But $H^1(\mathcal{O}(a-b_j))=0$ by Question 3. Adjusting the lifts by a coboundary makes them glue, splitting the sequence. This proves the Birkhoff–Grothendieck theorem here. The direct-sum Laurent calculation consequently proves the perfect pairingAlthough a splitting proves nondegeneracy, the pairing itself uses only evaluation followed by the coefficient trace on $H^1({\Omega }_{{\mathbb{P}}^1}^1)$, so does not depend on that splitting. This is residue duality on the projective line.

Now choose a rational function $t$ on $C$ with $dt\ne 0$. Such a choice is available over the perfect field $k$: a local parameter at a smooth point has nonzero differential in the rank-one sheaf of Kähler differentials over a field. Since $dt$ spans the differentials of $k(C)/k$, the extension $k(C)/k(t)$ has zero relative differentials and is separable. Question 1 gives a finite generically separable morphism $\pi :C\to {\mathbb{P}}^1$. It is flat: locally its finite module over a discrete valuation ring is torsion-free, hence free. For the invertible sheaf $\mathcal{M}$, its pushforward $E={\pi }_{\ast }\mathcal{M}$ is therefore a vector bundle.

We next prove the differential-valued dual identification needed to transfer the pairing. At a point of the base, complete the local rings. Each branch is of the formIf $e$ is the order of $f$, formal division writes $B=A[u]$ with basis $1,u,\dots ,u^{e-1}$ and a monic degree-$e$ equation $P(u)=0$. In particular this argument does not assume tame ramification or the special equation $s=u^e$. For a polynomial $g$ the elementary interpolation identity isTo see it, pass to a field containing the distinct roots $u_i$ of $P$. Lagrange interpolation writes the remainder as ${\sum }_{i}g(u_i)P(T)/((T-u_i)P^{\prime }(u_i))$. Comparison of its leading coefficient gives the identity, and descent returns it to the original field. The top-coefficient pairing on $A[T]/(P)$ is perfect: in the power basis its matrix has zero entries for $i+j<e-1$ and ones for $i+j=e-1$, so its determinant is a unit. It follows that the trace-dual fractional module is exactly $P^{\prime }(u{)}^{-1}B$.

Formal division of $f(T)-s$ gives $f(T)-s=W(T,s)P(T,s)$ with $W$ a unit. Differentiate in $T$ and evaluate at $u$. This shows that $f^{\prime }(u)$ and $P^{\prime }(u)$ differ by a unit. Since $ds=f^{\prime }(u)du$, we obtainThis is the trace-dual module of a finite curve map identity. Its map sends a differential $\eta$ to the functional $b\mapsto \mathrm{Tr}(b\eta )$, where the trace of $hds$ is $\mathrm{Tr}(h)ds$. For several branches take the direct sum of these maps. Completion is faithfully flat, so their being isomorphisms proves the corresponding isomorphism of the original local modules. A local trivialization of $\mathcal{M}$ now gives, compatibly on overlaps,

The inverse images of the two standard affine charts and their intersection are affine. As in Question 4, the Čech cochain complexes identify $H^1(C,\mathcal{M})$ with $H^1({\mathbb{P}}^1,E)$. Global sections in the last display identify the other factor with $H^0(C,{\mathcal{M}}^{\vee }\otimes {\Omega }_C^1)$. Thus the perfect pairing already calculated on the projective line givesIn particular the stated vanishing of global sections forces $H^1(C,\mathcal{M})=0$. The proof has established the relevant case of Serre duality, rather than invoked it as the requested argument.

To identify its intrinsic trace, use the rational principal-parts resolution on an algebraic curve. The constant sheaf of rational sections and the sheaf of their finite-support principal parts are both flasque sheaves. Consequently a class of $H^1(C,\mathcal{M})$ is represented by finitely many rational principal parts $(m_P)$, modulo the family coming from one global rational section. For a global section $\eta$ of ${\mathcal{M}}^{\vee }\otimes {\Omega }_C^1$, evaluation gives the pairingThe algebraic residue of a rational differential is its $u^{-1}du$ coefficient, independent of the local parameter. Adding a regular representative changes no residue. The local trace calculation above also givesOne can check this directly in the monogenic description: expand a differential in the basis $1,u,\dots ,u^{e-1}$, use the displayed interpolation formula for its trace, and extract the coefficient of $s^{-1}ds$. Formal division of $f(u)-s$ makes that coefficient exactly the coefficient of $u^{-1}du$; for a product of branch fields the trace is the sum. Thus the identity remains valid in positive characteristic, including wild ramification. On ${\mathbb{P}}^1$, partial fractions show that a rational differential's residues at finite points sum to the negative of its residue at infinity. Tracing to ${\mathbb{P}}^1$ therefore proves that the sum of residues of every global rational differential on $C$ is zero. Adding a global rational section to $(m_P)$ consequently changes the pairing by zero. This defines the same pairing as the finite-pushforward calculation, intrinsically, without a chosen map to ${\mathbb{P}}^1$.

Finally take $\mathcal{M}={\Omega }_C^1$. Its differential-valued dual is ${\mathcal{O}}_C$, and Question 4 proved $H^0(C,{\mathcal{O}}_C)=k$. Perfectness identifies $H^1(C,{\Omega }_C^1)$ with the dual of this space. Evaluation at the distinguished constant section $1$ givesThis residue trace on a smooth projective curve uses neither a basis choice nor an auxiliary finite map. It is the canonical isomorphism.