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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 $p$.

Let $F$ be globally generated of rank $r\ge 2$ on a projective curve, and put $W=H^0(C,F)$. The pairs $(p,s)$ with $s(p)=0$ form the total space of the kernel of the evaluation map over the curve. This incidence variety has dimension $\dim W+1-r<\dim W$. Its image in $W$ is closed because the curve is proper, and is a proper subset by the dimension bound. A section outside it vanishes nowhere and gives a line subbundle ${\mathcal{O}}_C\subset F$ with locally free quotient.

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.

We prove the Birkhoff–Grothendieck theorem by induction on the rank of the vector bundle $E$. Rank zero is immediate; rank one is the classification $\mathrm{Pic}({\mathbb{P}}_k^1)\cong \mathbb{Z}$ established above.

First there is a largest integer $a$ with $H^0(E(-a))\ne 0$. To see existence and boundedness directly, trivialize $E$ on the two standard affine charts: the associated finite projective modules over the principal ideal domains $k[t]$ and $k[t^{-1}]$ are free. Let $A(t)\in G{L}_r(k[t,t^{-1}])$ be its transition matrix, so a section of $E(-a)$ is a pair satisfyingIf $a$ exceeds the largest exponent in any entry of $A$, the right side has only strictly negative powers; the left side has nonnegative powers. Hence both vanish. For $a$ sufficiently negative, take a nonzero constant vector $p$ and put $q=t^{a}A(t{)}^{-1}p$; all its exponents are nonpositive, so it supplies a nonzero section. Thus a largest $a$ exists.

Such a section gives a nonzero map $\mathcal{O}(a)\to E$. It is injective, since it is nonzero generically and the source is torsion-free. Saturate its image. The resulting line subbundle is $\mathcal{O}(a+d)$ for the degree $d\ge 0$ of the effective divisor of zeros of the original map. If $d>0$, its inclusion in $E$ would give a nonzero section of $E(-a-d)$, contradicting maximality. Therefore $d=0$ andhas a locally free quotient. By induction, $Q\cong {⨁}_j\mathcal{O}(b_j)$.

Twist this exact sequence by $\mathcal{O}(-a-1)$. The outer groups $H^0(\mathcal{O}(-1))$ and $H^1(\mathcal{O}(-1))$ are zero by the Čech cohomology of twists on the projective line. The long exact sequence in sheaf cohomology therefore givesThe first is zero by the choice of $a$, so each $H^0(\mathcal{O}(b_j-a-1))$ vanishes. Thus $b_j\le a$ for every $j$. The extension class lies insince $a-b_j\ge 0$. The identification of this Ext functor with cohomology can be seen without additional duality: local splittings of a vector-bundle extension differ by a Čech cocycle in $\mathcal{H}om(Q,\mathcal{O}(a))$, and a coboundary changes them to compatible splittings. Vanishing therefore gives a global splitting. Consequentlycompleting the induction over any field $k$.

The degrees can be arranged in nonincreasing order and their multiset is unique. Indeed, for any such decomposition,so the dimensions of the twisted global sections recover the number of summands of every degree.

Use the finite map $f:C\to {\mathbb{P}}^1$ from the preceding solution. The inverse images of the two standard affine charts and of their intersection are affine, because $f$ is finite. The vanishing of quasi-coherent cohomology on an affine scheme therefore lets their two-term Čech cochain complex compute $H^i(C,E)$. It has terms only in degrees zero and one, provingThis actually applies to every quasi-coherent sheaf on $C$.

For a smooth projective curve with $H^0(C,{\mathcal{O}}_C)=k$, define its genus of a smooth projective curve byThe preceding Serre duality also gives $g={\dim }_k{H}^0(C,{\Omega }_C^1)$. The condition on constants holds in particular for a geometrically connected smooth projective curve. It makes $g$ equal to the arithmetic genus $1-\chi ({\mathcal{O}}_C)$. We state the theorem with this usual convention and then give the formula valid without that condition.

For a divisor on an algebraic curve $D$ and a canonical divisor $K$ with $\mathcal{O}(K)\cong {\Omega }_C^1$, the Riemann-Roch theorem isThe degree over an arbitrary field is weighted by residue degrees: ${\mathrm{deg}}_{k}D={\sum }_p{n}_p[k(p):k]$ when $D={\sum }_p{n}_{p}p$. Equivalently, for a line bundle $L$,We now prove it via Riemann-Roch via elementary modifications.

For any closed point $p$ and any line bundle $L$, the exact sequencehas a skyscraper sheaf as its quotient, with $k$-dimension $[k(p):k]$. This follows locally from a uniformizer of the discrete valuation ring at $p$: a local line-bundle generator modulo that uniformizer spans one copy of $k(p)$. The quotient has zero higher cohomology, as it is the direct image from a finite affine scheme. The long exact sequence in sheaf cohomology and the proved higher vanishing giveEvery line bundle has a nonzero rational section of a line bundle, whose local orders give a divisor on an algebraic curve $D$ with $L\cong \mathcal{O}(D)$. Iterating the displayed identity, adding or subtracting closed points with their multiplicities, givesFor the constants convention above, $\chi ({\mathcal{O}}_C)=1-g$. The dualizing sheaf on a smooth projective curve gives $h^1(\mathcal{O}(D))=h^0(\mathcal{O}(K-D))$, which converts this Euler-characteristic identity into the stated Riemann-Roch theorem. This proves the formula over arbitrary $k$, without assuming the closed points are $k$-rational.

For vector bundles, the corresponding formula isTo see it, saturate a rational rank-one subspace to a line subbundle and induct on rank. Both Euler characteristic of a coherent sheaf and the degree of the determinant are additive in the resulting short exact sequence. The line-bundle formula just proved starts the induction.

Taking $L={\Omega }_C^1$ and using duality gives $h^0({\Omega }_C^1)=g$, $h^1({\Omega }_C^1)=1$, and hence ${\mathrm{deg}}_k{\Omega }_C^1=2g-2$. In particular a line bundle of degree greater than $2g-2$ has zero first cohomology: its dual twisted by ${\Omega }_C^1$ has negative degree, and a negative-degree line bundle has no nonzero section, since any section would have an effective divisor of zeros.

If the term “curve over $k$” is used without requiring $H^0({\mathcal{O}}_C)=k$, the unconditional formula we proved isWriting $g=h^1({\mathcal{O}}_C)$ then replaces $1-g$ by $h^0({\mathcal{O}}_C)-g$. For example, ${\mathbb{P}}_{k^{\prime }}^1$ viewed as a $k$-curve for a finite separable extension $k^{\prime }/k$ has $h^0({\mathcal{O}}_C)=[k^{\prime }:k]$, so a literal constant term $1$ would be incorrect. Alternatively the arithmetic genus $p_a=1-\chi ({\mathcal{O}}_C)$ gives the formula with constant $1-p_a$. The constants hypothesis for the usual genus is recorded in Stacks Project, genus of a curve.

The coefficient of $t^{-1}dt$ gives a trace on $H^1({\mathbb{P}}^1,{\Omega }^1)$. Multiplication followed by this trace pairs the Laurent bases of the Čech cohomology of twists on the projective line perfectly: $t^a$ pairs with $t^{-a-1}$. For any vector bundle, evaluation into ${\Omega }^1$ defines the same natural pairing. Saturate a rational rank-one subspace to a line subbundle, and induct on rank using the long exact sequence in sheaf cohomology and compatibility of evaluation with connecting maps. This proves the duality without using the later splitting theorem.

For a rank-two bundle $E$, the sufficiently positive twist $E(m{p}_0)$ is globally generated. The nowhere-vanishing section of a globally generated bundle on a curve gives a subbundle ${\mathcal{O}}_C\subset E(m{p}_0)$. Twisting back produces a genuine line subbundle of degree $-m$, with locally free quotient. Thus line-subbundle degrees are bounded above but not below. Merely inserting a negative divisor on an algebraic curve into a fixed subline would allow a torsion quotient and would not establish this stronger conclusion.

Choose a rational basis of the dual of a vector bundle and an effective divisor $D$ bounding all its poles. Evaluation gives a generically injective map $E\to {\mathcal{O}}_C(D{)}^{\oplus r}$. A line subbundle maps nontrivially to at least one summand, so ${\mathcal{O}}_C(D)\otimes L^{-1}$ has a nonzero regular section. Consequently $\mathrm{deg}L\le \mathrm{deg}D$. This bound depends on $E$ and the chosen rational frame, and does not require $E$ to be stable.