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Over any field, the canonical bundle of projective space is $\mathcal{O}(-n-1)$. Indeed, dualizing the Euler sequence gives $0\to {\Omega }_{{\mathbf{P}}^n}^1\to \mathcal{O}(-1{)}^{\oplus (n+1)}\to \mathcal{O}\to 0$; taking determinants gives the result. Hence Serre duality identifies $H^i({\mathbf{P}}^n,\mathcal{O}(m))$ with the dual of $H^{n-i}({\mathbf{P}}^n,\mathcal{O}(-m-n-1))$.

On a smooth projective curve, the exact sequence $0\to \mathcal{O}(D-P)\to \mathcal{O}(D)\to k(P)\to 0$ increases the Euler characteristic by one. Iteration for positive and negative coefficients gives $\chi (\mathcal{O}(D))=\mathrm{deg}D+1-g$. Serre duality identifies $h^1(\mathcal{O}(D))$ with $h^0(\mathcal{O}(K-D))$, proving $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$.

For every integer $d$, the cohomology of a twisting sheaf on projective space isThus all intermediate cohomology vanishes. The top line follows from Serre duality and ${\omega }_{{\mathbb{P}}^n}\cong \mathcal{O}(-n-1)$.

For an effective divisor on an algebraic curve $D$ on a smooth projective curve, the Zariski tangent space to its symmetric product of a curve is $H^0({\mathcal{O}}_D(D))$. The derivative of the Abelian sum map is the connecting homomorphism of $0\to {\mathcal{O}}_X\to {\mathcal{O}}_X(D)\to {\mathcal{O}}_D(D)\to 0$. Its dual under Serre duality is restriction of holomorphic differential forms to $D$, including all jets at repeated points. Hence its rank is $\mathrm{deg}D+1-h^0({\mathcal{O}}_X(D))$, and its image annihilates $H^0(K_X(-D))$.

Choose an effective Cartier divisor $E$ of degree at least $g$ on a smooth projective genus-$g$ algebraic curve. For a family $M$ of degree-$g-1$ line bundles, the evaluation map from ${\pi }_{\ast }M(E)$ to ${\pi }_{\ast }(M(E){|}_E)$ is between vector bundles of the same rank. Its determinant vanishes exactly when $H^0(C,M)$ is nonzero. Changing $E$ changes the presentation by cancelling an invertible determinant factor; the zero Cartier divisor is the theta divisor. At a point with $h^0(M)=h^1(M)=1$, the derivative is the cup-product map $H^1({\mathcal{O}}_C)\to \mathrm{Hom}(H^0(M),H^1(M))$. By Serre duality it is nonzero, since the product of nonzero sections of $M$ and $K_C\otimes M^{-1}$ is nonzero. Hence this Cartier divisor is generically reduced.

On ${\mathbb{P}}_k^n$ with $n>0$, let $\mathcal{F}=i_{\ast }k$ at a rational point. It has one-dimensional global sections, but its dual of a sheaf is zero: a local map from the residue field into the local polynomial ring must have image annihilated by a nonzero parameter, which is impossible in an integral domain. Thus $H^n({\mathcal{F}}^{\vee }(-n-1))=0$. Serre duality for arbitrary coherent sheaves instead uses an Ext functor; replacing this by the ordinary dual can lose torsion information.

If $D$ is a nontrivial nef divisor on a K3 surface with $D^2=0$, its complete linear system of a divisor has no fixed part. Write $D=F+M$. Nefness of $D$ and $M$ gives $D\cdot F=D\cdot M=M^2=M\cdot F=F^2=0$. But a nonzero fixed part satisfies $h^0(F)=1$, while Riemann–Roch theorem for algebraic surfaces and Serre duality would give $h^0(F)\ge 2$ if $F^2=0$. Hence $F=0$. Two movable members with no common component have intersection zero and are disjoint, so the system is basepoint-free and has Iitaka dimension one.

For a smooth projective curve with $H^0(C,{\mathcal{O}}_C)=k$, its genus is ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$, equivalently ${\dim }_k{H}^0(C,{\Omega }_C^1)$ by Serre duality. It equals the arithmetic genus $1-\chi ({\mathcal{O}}_C)$. Without the condition on constants, the general Riemann-Roch theorem instead uses $\chi ({\mathcal{O}}_C)=h^0({\mathcal{O}}_C)-h^1({\mathcal{O}}_C)$.

For a vector bundle of finite positive rank let $B$ be an upper degree bound for the line subbundles of $E^{\ast }\otimes K_C$. If $m-1>B$, no nonzero map ${\mathcal{O}}_C(m{p}_0-p)\to E^{\ast }\otimes K_C$ can exist for any point $p$: its saturated image would have degree at least $m-1$. Serre duality then gives $H^1(E(m{p}_0-p))=0$. The evaluation exact sequence shows that $H^0(E(m{p}_0))\to E(m{p}_0){|}_p$ is surjective at every point, proving the assertion with a uniform bound in $p$.

Suppose a line bundle $L$ of degree $g-1$ on a smooth projective curve has $h^0(L)=h^1(L)=r$. Choose $r$ distinct points where evaluation is invertible for both $H^0(L)$ and $H^0(K_X\otimes L^{-1})$. Such choices exist simultaneously: each determinant defines a nonempty open subset of $X^r$, since independent sections cannot vanish at every point. Let $\xi \in H^1({\mathcal{O}}_X)=H^0(K_X{)}^{\ast }$ be a sum of nonzero weighted evaluation functionals at these points. The Serre duality matrix of $\xi ⌣-$ factors as the product of the two invertible evaluation matrices and an invertible diagonal matrix. Its determinant is nonzero. This gives the nonzero leading determinant needed for the Riemann singularity theorem.

Let $D={\sum }_p{r}_{p}p\in X^{(d)}$. The derivative of the Abelian sum map is most naturally stated using principal parts. The tangent space of the symmetric product of a curve at $D$ is $H^0({\mathcal{O}}_D(D))$, and the tangent space of the Jacobian variety is $H^0(K_X{)}^{\ast }$. With these identifications,Its dual is restriction of holomorphic one-forms to the length-$d$ divisor scheme. In particular,The last equality is the Riemann-Roch theorem. These formulas apply to repeated points as well as distinct ones.

Here is a direct proof of the repeated-point Abel differential formula. At a multiplicity-$r$ point, choose a local coordinate $t$ with $t(p)=0$. A nearby local divisor is described by the roots $t_1,\dots ,t_r$, or, without making an ordering choice, by their elementary symmetric functions $e_1,\dots ,e_r$:These $e_k$ are smooth local coordinates on the symmetric product of a curve, including at the repeated divisor. Under a first-order deformation, minus the logarithmic variation of this defining polynomial is the principal partThe sign can be checked at $r=1$: moving the point from $0$ to $ϵa$ gives $v=a/t$, and the Abel integral changes by $a\omega (p)$.

Write $\omega =({\sum }_{j\ge 0}{c}_j{t}^j)dt$ and its local primitive as $F(t)={\sum }_{j\ge 0}{c}_j{t}^{j+1}/(j+1)$. The local contribution to the Abelian sum map is ${\sum }_{i}F(t_i)$. Newton's identities, linearized at $e_1=\cdots =e_r=0$, giveSubstituting in the primitive therefore givesSumming over the support proves the theorem. In particular, a multiplicity-$r$ point tests the first $r$ coefficients of a holomorphic one-form, not just its value. The residue pairing between principal parts of order at most $r$ and differential jets through order $r-1$ is perfect. Thus the annihilator of the derivative is precisely the holomorphic one-forms vanishing to order at least $r_p$ at every $p$, namely $H^0(K_X(-D))$.

There is also a sheaf cohomology description that explains its kernel geometrically. The sequencegives a connecting homomorphism $\delta :H^0({\mathcal{O}}_D(D))\to H^1({\mathcal{O}}_X)$. Represent a principal part by meromorphic lifts on coordinate disks; their differences on overlaps give its Čech cohomology class. Under Serre duality, pairing this class with $\omega$ is the sum of the residues of $v_p\omega$, so $\delta$ is the displayed derivative. Exactness shows that its kernel is $H^0({\mathcal{O}}_X(D))/\mathbb{C}$, of dimension $h^0({\mathcal{O}}_X(D))-1$, in agreement with the tangent directions to the complete linear system of a divisor in Solution 2.

Two consequences will be useful below. If $d\ge 2g-1$, then $K_X(-D)$ has negative degree and no nonzero sections, so $u_d$ is a submersion everywhere. Its image is open, and it is also closed because $X^{(d)}$ is compact. Since the Jacobian variety is connected, $u_d$ is surjective. Thus the points $u(p)$ generate the Jacobian variety as a group. Together with Abel's theorem, this identifies the degree-zero Picard group with the Jacobian variety: every line bundle has a meromorphic section and thus a divisor representation, injectivity is Abel's theorem, and surjectivity follows from $u_d(D)-du(p_0)$.

For $g\ge 1$, one can choose $g-1$ distinct points imposing independent conditions on $H^0(K_X)$: at each step a surviving nonzero holomorphic one-form has only finitely many zeros, so choose a point where it does not vanish. Their sum $D$ has $h^0(K_X(-D))=1$ and hence $h^0({\mathcal{O}}_X(D))=1$. The derivative has rank $g-1$ there. Consequently $W_{g-1}=u_{g-1}(X^{(g-1)})$ is an irreducible subvariety of dimension $g-1$. This also covers $g=1$, when $X^{(0)}$ and $W_0$ are points.

Identify ${\mathrm{Pic}}^{g-1}(X)$ with the Jacobian variety using the base point and the translation of Solution 4. A line bundle $L$ of degree $g-1$ corresponds to the point $z=u(D)-\kappa$ for any divisor representation $L\simeq {\mathcal{O}}_X(D)$. The Riemann singularity theorem isIn particular, the theta divisor is singular exactly at the degree-$g-1$ line bundles with at least two independent sections. We prove the equality, including the upper bound on multiplicity.

Choose a fixed effective divisor $A$ of degree $N\ge g$, and a local holomorphic family $L_{\eta }$ representing nearby points of ${\mathrm{Pic}}^{g-1}(X)$. Such a family can be constructed locally from the holomorphic exponential sequence, by varying line-bundle transition functions through exponentials of Čech cohomology representatives. By Serre duality, $H^1(L_{\eta }(A))$ is dual to $H^0(K_X\otimes L_{\eta }^{-1}(-A))$, whose degree is $g-1-N<0$, so it vanishes. The Riemann-Roch theorem gives $h^0(L_{\eta }(A))=N$. By cohomology and base change for line bundles on a curve, these spaces form a holomorphic vector bundle $E$ of rank $N$. The spaces $H^0(L_{\eta }(A){|}_A)$ form another holomorphic vector bundle $F$ of rank $N$. Evaluation gives a holomorphic square matrix $T(\eta )$ locally, with exact sequenceConsequently $\det T(\eta )=0$ exactly when $L_{\eta }$ has a nonzero section, namely on the support of the theta divisor.

At the point $L$, put $r=h^0(L)=h^1(L)$, the equality being Riemann-Roch theorem. An invertible minor of size $N-r$ persists near this point. Holomorphic row and column operations, or the Schur complement, givewhere $M$ is an $r\times r$ holomorphic matrix. This is the Schur complement presentation of a theta singularity. Its linear term in a tangent direction $\xi \in H^1({\mathcal{O}}_X)$ is the cup product mapTo see this concretely, vary transition functions $g_{ij}$ of $L$ to $g_{ij}(1+ϵ{\xi }_{ij})$. A section $s_i$ lifts to $s_i+ϵs_i^{\prime }$ precisely when its first-order compatibility equation can be solved; the obstruction is the cocycle ${\xi }_{ij}{s}_j$. This is $\xi ⌣s$. Passing to the kernel and cokernel of $T(0)$ gives exactly the derivative of the reduced matrix $M$, proving the assertion without assuming a multiplicity formula.

Before extracting the order of $\det M$, we verify that $\det T$ is a reduced equation of the theta divisor. Generic points of $W_{g-1}$ have $h^0(L)=1$ by Solution 3. At such a point, both $H^0(L)$ and $H^0(K_X\otimes L^{-1})$ are one-dimensional. Their nonzero sections $s,t$ have nonzero product $st\in H^0(K_X)$. Choose $\xi \in H^1({\mathcal{O}}_X)$ with $⟨\xi ,st⟩\ne 0$ under Serre duality. The scalar $d{M}_0(\xi )$ is nonzero, so $\det T$ has a simple zero generically on the irreducible theta divisor. Since its support is exactly that irreducible hypersurface, it defines the reduced divisor. Solution 4 independently proved the same reduced structure for the Riemann theta function. On the smooth ambient Jacobian variety, local equations of the same reduced divisor differ by a nonvanishing holomorphic unit. Hence the orders of the local theta equation and $\det M$ agree everywhere.

Every entry of $M$ vanishes at the point, so each term in its determinant has order at least $r$. This proves ${\mathrm{mult}}_z\Theta \ge r$. To obtain equality, we exhibit an invertible cup-product direction for a line bundle on a curve. Set $V=H^0(L)$ and $W=H^0(K_X\otimes L^{-1})$, both of dimension $r$. There are $r$ distinct points $p_1,\dots ,p_r$ at which evaluation is an isomorphism for $V$: choose them inductively, since a surviving nonzero section has only finitely many zeros. The same holds for $W$. The nonvanishing of each evaluation determinant defines a nonempty open subset of the irreducible variety $X^r$. Their intersection, also avoiding the diagonals, is therefore nonempty. Choose points in it and local coordinates and compatible trivializations of $L$ and $K_X\otimes L^{-1}$ at each point.

Let $\xi$ be the connecting homomorphism class of the meromorphic principal parts ${\lambda }_i/t_i$ at these points, with all ${\lambda }_i\ne 0$. The residue form of Serre duality givesIf $S$ and $U$ are the two invertible evaluation matrices, the matrix of this pairing is $S^t\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_r)U$, and is invertible. Thus the cup product map $\xi ⌣-:H^0(L)\to H^1(L)$ is an isomorphism. Along the corresponding analytic parameter line,with nonzero leading coefficient. The order is at most $r$, completing the proof of the Riemann singularity theorem.

This argument also identifies the tangent cone to a theta divisor: its degree-$r$ equation is $\det (\xi ⌣-)$, which is not the zero polynomial. For $r=1$, the tangent hyperplane is $⟨\xi ,st⟩=0$, in agreement with the annihilator description of the derivative of the Abelian sum map. For $r\ge 2$ all first derivatives vanish, proving singularity. In genus $1$ the theta divisor is a single smooth point of multiplicity $1$. In genus $0$ the effective degree-$-1$ locus is empty, so there is no singularity statement at a point of it.

Let $X$ be a separated Noetherian scheme over a field $k$ and $\mathcal{F}$ a coherent sheaf. Coherent cohomology studies the groups $H^q(X,\mathcal{F})$: global sections in degree zero and the obstructions to gluing local sections in higher degrees. Coherence means finite generation locally, and on a Noetherian scheme it is preserved by kernels, cokernels and extensions. This makes coherent sheaves suitable for exact-sequence calculations.

For an ordered open cover $\mathcal{U}=(U_i)$, the Čech cochain groups arewith differentialEach double omission occurs twice with opposite signs, so ${\delta }^2=0$. The Čech cohomology of the cover is ${\check{H}}^q=\ker \delta /\mathrm{im}\delta$, and taking the direct limit over refinements defines cover-independent Čech cohomology. In degree zero the sheaf gluing axiom identifies it with $\Gamma (X,\mathcal{F})$.

The comparison with derived-functor sheaf cohomology is particularly effective algebraically. Positive-degree cohomology of a quasi-coherent sheaf on an affine scheme vanishes. A separated scheme has affine intersections of affine opens, so a finite affine cover is acyclic for $\mathcal{F}$. Its Čech cochain complex consequently computes $H^q(X,\mathcal{F})$ directly. For a short exact sequence of sheaves, sections over each affine intersection are exact on quasi-coherent sheaves. The resulting short exact sequence of Čech complexes gives the long exact sequence in sheaf cohomology. In particular, the connecting map measures the failure of a global section of the quotient to lift globally.

Several central projective results make the theory finite and computable. On a projective variety over $k$, every coherent cohomology group is finite-dimensional. The Grothendieck vanishing theorem gives $H^q(X,\mathcal{F})=0$ for $q>\dim X$. If ${\mathcal{O}}_X(1)$ is ample, Serre vanishing gives $H^{q>0}(X,\mathcal{F}(m))=0$ for sufficiently large $m$, and these large twists are globally generated sheaves. On projective space a coherent sheaf has a finite twisting resolution of a coherent sheaf on projective space: finite sums of $\mathcal{O}(a)$ resolve it. The explicit line-bundle cohomology below and the long exact sequences prove finiteness and positive-twist vanishing. For a closed embedding $i:X\hookrightarrow {\mathbb{P}}^n$, $i_{\ast }\mathcal{F}$ is coherent and $H^q(X,\mathcal{F})=H^q({\mathbb{P}}^n,i_{\ast }\mathcal{F})$, reducing these results to projective space. A sufficiently positive twist of a finite presentation by sums of $\mathcal{O}(-a)$ also proves global generation.

The Euler characteristic of a coherent sheafis therefore defined, and the long exact sequence makes it additive. With a fixed projective embedding, $\chi (X,\mathcal{F}(m))$ is its Hilbert polynomial, and for sufficiently large $m$ it equals $h^0(X,\mathcal{F}(m))$. On a smooth projective variety of dimension $r$, Serre duality gives, for a vector bundle $E$,For an arbitrary coherent sheaf the correct counterpart is $H^q(X,\mathcal{F}{)}^{\ast }\cong {\mathrm{Ext}}_X^{r-q}(\mathcal{F},{\omega }_X)$; replacing this Ext group indiscriminately by the cohomology of the ordinary sheaf dual would lose local extension information.

We now calculate the basic examples. For $n\ge 1$, use the standard cover $U_i=D_{+}(x_i)$ of ${\mathbb{P}}^n$. Sections of $\mathcal{O}(m)$ on an intersection are degree-$m$ Laurent monomials in which negative exponents are allowed only for coordinates inverted on that intersection. Decompose the Čech cochain complex by exponent vectors $(a_0,\dots ,a_n)$ with sum $m$. If no exponent is negative, the corresponding simplex complex contributes only in degree zero. If all exponents are negative, the monomial occurs only on the full intersection and contributes only in degree $n$. When the negative-index set is nonempty and proper, inserting any index outside it gives a contracting homotopy, so the monomial complex is acyclic. Hence the cohomology of twisting sheaves on projective space isIn the top group a basis consists of the Laurent monomials with every exponent at most $-1$. Writing $a_i=-1-b_i$ reduces the count to nonnegative solutions of $\sum b_i=-m-n-1$, giving the displayed binomial coefficient. In particular, all intermediate cohomology vanishes, and ${\omega }_{{\mathbb{P}}^n}=\mathcal{O}(-n-1)$ agrees with Serre duality. For ${\mathbb{P}}^0$ separately, every twist is trivial and $H^0=k$.

On the projective line this givesThus positive and negative degrees produce quite different groups, while their alternating difference is a single polynomial. The twisted cubic is isomorphic to ${\mathbb{P}}^1$ and its hyperplane bundle corresponds to ${\mathcal{O}}_{{\mathbb{P}}^1}(3)$. Consequently $h^0(X,{\mathcal{O}}_X(m))=3m+1$ for $m\ge 0$, agreeing with Q2, and $h^1(X,{\mathcal{O}}_X(m))=\max (-3m-1,0)$.

For a smooth plane curve $C$ of degree $d\ge 1$, the ideal-sheaf sequence of a projective hypersurface isThe long exact sequence in sheaf cohomology and the preceding formula give $H^0(C,{\mathcal{O}}_C)=k$ and $H^1(C,{\mathcal{O}}_C)\cong H^2({\mathbb{P}}^2,\mathcal{O}(-d))$. ThereforeFor $d=1,2$ this is zero; for a plane cubic it is one. Twisting the same exact sequence gives $\chi (C,{\mathcal{O}}_C(m))=dm+1-g$. These examples illustrate how Čech cohomology, vanishing, Serre duality and exact sequences convert local equations into global numerical invariants.

Let $C$ be the smooth algebraic curve. A nonzero regular section of a line bundle has an effective zero divisor, so its degree of a line bundle is nonnegative. Degrees add under tensor products and change sign on taking the dual. We use genuine line subbundles, so their quotients in $E$ are locally free. Both bounds can be proved constructively.

For the upper bound, take a rational frame of $E^{\ast }$ over the function field of $C$. Its two members are rational sections with only finitely many poles. Choose an effective divisor $D$ large enough to bound the pole orders of both rational sections. They become global sections of $E^{\ast }\otimes {\mathcal{O}}_C(D)$ and define a sheaf mapwhich is an isomorphism at the generic point. It is therefore injective as a sheaf map: a section in its kernel would vanish generically in a locally free sheaf, which is a torsion-free sheaf. For any line subbundle $L\subset E$, at least one of its two component maps into ${\mathcal{O}}_C(D)$ is nonzero. Such a map is a nonzero global section of ${\mathcal{O}}_C(D)\otimes L^{-1}$. Its zero divisor on an algebraic curve is effective, soThis upper degree bound for line subbundles on a smooth curve depends only on a choice made for $E$, and applies to every one of its line subbundles.

For the lower-bound question we will actually construct subbundles ${\mathcal{O}}_C(-m{p}_0)$ for every sufficiently large integer $m$, where $p_0$ is a fixed point. First prove the globally generated vector bundle property rather than assuming that negative-degree inclusions automatically are subbundles.

Apply the upper-bound argument to the rank-two vector bundle $F=E^{\ast }\otimes K_C$, where $K_C$ is the canonical bundle. Let $B$ be its upper degree bound. Choose $m$ with $m-1>B$. For any point $p\in C$, a nonzero global section of $F(-m{p}_0+p)$ would give a nonzero mapIts image can be saturated to a saturated line subbundle on a smooth curve $M\subset F$. Locally the local ring of a smooth algebraic curve is a discrete valuation ring: dividing out the common power of a uniformizer makes the image a primitive rank-one summand, and its quotient is free. Thus the saturation is indeed a subbundle. The original map gives a nonzero global section of $M\otimes {\mathcal{O}}_C(-m{p}_0+p)$ and consequently $\mathrm{deg}M\ge m-1$, contradicting the upper bound. Hence $H^0(F(-m{p}_0+p))=0$ for every $p$. By Serre duality,The exact sequenceand its long exact sequence in sheaf cohomology show that evaluation $H^0(E(m{p}_0))\to E(m{p}_0){|}_p$ is onto for every $p$. This proves the global generation of positive twists on a smooth curve with one bound on $m$ that works at every point.

Write $W=H^0(E(m{p}_0))$ and $N=\dim W$. The evaluation map $W\otimes {\mathcal{O}}_C\to E(m{p}_0)$ is surjective with kernel bundle of rank $N-2$. Consider the incidence subsetIt is the total space of this kernel bundle, so $\dim Z=1+(N-2)=N-1$. Its projection into $W$ is closed because $C$ is proper, and its image has dimension at most $N-1<N$. Choose $s$ outside this proper closed subset. It vanishes nowhere. In a local trivialization at any point, one component of $s$ is a unit, so it can be completed to a basis. Therefore ${\mathcal{O}}_C\stackrel{s}{\to }E(m{p}_0)$ is a subbundle with a locally free rank-one quotient. Twisting back givesThis proves the unbounded negative degrees of line subbundles. The nowhere-vanishing construction guarantees a subbundle at every point, including the zeros that might otherwise have occurred in a generic meromorphic inclusion. If the curve has several connected components, apply these constructions on each component and add the degrees; the same upper-bound and no-lower-bound conclusions follow.

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.

Let $S=k[X_0,\dots ,X_n]$ with its usual grading, and put $S(m{)}_d=S_{m+d}$. Define the twisting sheaf on projective space byMore explicitly, on $U_i=D_{+}(X_i)$ its sections are the homogeneous degree-$m$ part of $S_{X_i}$. They are a free rank-one module over $(S_{X_i}{)}_0=k[X_j/X_i:j\ne i]$, with generator $e_i=X_i^m$. These generators make sense for every integer $m$, since $X_i$ is invertible on this chart. Their transition rule isThe ratios are regular units on overlaps and satisfy the cocycle identity, so they glue to an invertible sheaf. In particular $\mathcal{O}(m)\otimes \mathcal{O}(r)\cong \mathcal{O}(m+r)$.

For $n\ge 1$, a global section is a compatible homogeneous degree-$m$ rational expression lying in every $S_{X_i}$. Inside the fraction field of $S$,To justify this, write a fraction in lowest terms in the unique factorization domain $S$. Membership in $S_{X_i}$ forces every irreducible divisor of its denominator to be associated with $X_i$. Since there are at least two nonassociated variables, the denominator must be a unit. ThereforeThe degree-$m$ monomials are indexed by nonnegative integers $a_0,\dots ,a_n$ with sum $m$. The stars and bars count givesHere and below, an all-integer binomial shorthand must use the counting convention $\left(\genfrac{}{}{0ex}{}{a}{n}\right)=0$ when $a<n$, including negative $a$. The polynomial extension of binomial coefficients to negative upper arguments does not give the stated dimensions. In dimension zero, ${\mathbb{P}}^0$ is one point and every ${\mathcal{O}}_{{\mathbb{P}}^0}(m)$ is trivial, so its space of global sections is $k$ for every $m$; this case is treated separately.

For a closed inclusion $i:H\hookrightarrow {\mathbb{P}}^n$, direct image sheaf formation is exact: its stalk is the original stalk at a point of $H$ and zero outside $H$. It preserves flabby sheaves, since restrictions on the target are restrictions between the corresponding open subsets of $H$. Push forward a flasque resolution of $\mathcal{F}$; exactness gives a flasque resolution of $i_{\ast }\mathcal{F}$, and its complex of global sections is the original one becauseTaking cohomology of these identical complexes proves sheaf cohomology under a closed inclusion:This argument applies even without quasi-coherence.

We derive the remaining sheaf cohomology dimensions using the given eventual vanishing, the long exact sequence in sheaf cohomology, and induction on dimension. Multiplication by the linear equation of a hyperplane $H\cong {\mathbb{P}}^{n-1}$ gives the hyperplane exact sequence for twisting sheavesExactness can be checked in each chart by quotienting its polynomial ring by the equation of $H$; multiplication is injective because that equation is a non-zero-divisor. We use the closed-inclusion cohomology identification to replace the cohomology of the final sheaf by that on $H$.

The dimension-zero base is a point, whose global-section functor is exact. Thus all its higher sheaf cohomology is zero. For $n=1$, the relevant long exact sequence isWhen $m\ge 0$, restriction of homogeneous polynomials to the hyperplane point is surjective: after choosing that point as $[1:0]$, the section $X_0^m$ restricts to a generator. Therefore the consecutive $H^1$ groups are isomorphic for $m\ge 0$. The given vanishing for large $m$ propagates down to $m=-1$. For $m<0$, both displayed $H^0$ groups are zero, and the sequence becomesStarting with $H^1(\mathcal{O}(-1))=0$, induction downward gives $h^1(\mathcal{O}(m))=-m-1$ for $m\le -2$. For degrees $i>1$, the point contributes no cohomology to the exact sequence, so consecutive $H^i$ groups are isomorphic for every $m$; the given eventual vanishing makes them all zero. Hence the complete positive-degree formula holds for ${\mathbb{P}}^1$.

Now let $n\ge 2$ and assume the formulas for $H\cong {\mathbb{P}}^{n-1}$. For every $m$, the restrictionis surjective: for $m\ge 0$ extend a polynomial in the hyperplane coordinates to the same polynomial on ${\mathbb{P}}^n$, and for $m<0$ both groups vanish because $H$ has positive dimension. Consequently the exact sequence gives an injectionFor $2\le i\le n-1$, the induction hypothesis says $H^{i-1}(H,{\mathcal{O}}_H(m))=0$, and gives the same injection in degree $i$. For each fixed $m$, composing finitely many injections reaches a sufficiently large twist, whose higher cohomology is zero by the given hypothesis. ThereforeFor $i>n$, the two neighbouring hyperplane groups $H^{i-1}(H,{\mathcal{O}}_H(m))$ and $H^i(H,{\mathcal{O}}_H(m))$ are zero by induction. Consecutive $H^i$ groups on ${\mathbb{P}}^n$ are then isomorphic, and eventual vanishing again makes them zero. This derives vanishing above the dimension too, without needing a special projective-space cohomology theorem.