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For a smooth projective curve, the degree-zero Picard group consists of line bundles of degree zero up to isomorphism, with tensor product as addition. A nonzero rational section of a line bundle identifies each line bundle with $\mathcal{O}(D)$ for a divisor on an algebraic curve; degree is the sum of multiplicities weighted by residue-field degrees. Over an algebraically closed field, every degree-zero divisor is a sum of differences of points.

On an integral scheme, for a Cartier divisor locally represented by $a_i\in K^{\ast }$, its divisor line bundle is the subsheaf of rational functions locally equal to $a_i^{-1}{\mathcal{O}}_X$. Unit ratios glue these free rank-one modules. Given a nonzero rational section of a line bundle $s=a_i{e}_i$, the maps $e_i\mapsto a_i^{-1}$ identify that invertible sheaf with ${\mathcal{O}}_X(D)$. Replacing $s$ by a nonzero rational multiple changes $D$ by a principal Cartier divisor.

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.

First use the algebraically closed field convention. Every line bundle on a smooth projective curve has a nonzero rational section of a line bundle: trivialize it on a nonempty open set and take its generic generator. The orders of its zeros and poles give a finite divisor $D$, and the local trivializations identify the line bundle with $\mathcal{O}(D)$. For a degree-zero line bundle writesince every closed point is a $k$-point. ThereforeOnly finitely many factors occur, and negative tensor powers mean dual line bundles. This provesFor genus zero the group is trivial, so the conclusion remains valid.

For a general field, a closed point need not be rational, and its contribution to degree is weighted by its residue degree. Consequently rational points need not generate the degree-zero Picard group of a curve. Here is an explicit counterexample to that stronger reading. Let $C$ be the smooth projective curve over ${\mathbb{F}}_2$ obtained by completing the affine equationThe polynomial is irreducible over ${\overline{\mathbb{F}}}_2(x)$: a function $r^2+r$ has even pole order at infinity when it has a pole there, while the right side has pole order five. Its affine model is smooth because the partial derivative with respect to $y$ is one. The normalized projective completion is therefore a geometrically integral smooth projective curve.

There is exactly one point $p_{\infty }$ above infinity, and it is rational. Indeed the map $x:C\to {\mathbb{P}}^1$ has degree two. At a point above infinity put $v(x)=-e$, where the ramification index $e$ is at most two. The equation forces $v(y)<0$ and $2v(y)=5v(x)=-5e$. Hence $e=2$ and $v(y)=-5$. The degree formula then leaves exactly one point of residue degree one. Thus $x$ and $y$ have pole orders two and five there. There are no affine rational points: for either $x\in {\mathbb{F}}_2$, the right side is one, while $y^2+y=0$ for every $y\in {\mathbb{F}}_2$. Hence $C({\mathbb{F}}_2)=\{p_{\infty }\}$ and the image of its Abel map of a pointed smooth projective curve is trivial.

The equations $y=0$, $x^2+x+1=0$ define a degree-two closed point $D$, since $x^5+x+1$ vanishes modulo $x^2+x+1$. The divisor $D-2p_{\infty }$ has degree zero but is not principal. To prove this, a function with poles bounded by $2p_{\infty }$ is regular on the affine curve and hence has a unique expression $A(x)+B(x)y$. The possible pole orders of its two terms are $2\mathrm{deg}A$ and $2\mathrm{deg}B+5$, of opposite parity, so they cannot cancel. The pole bound forces $B=0$ and $\mathrm{deg}A\le 1$. But no nonzero $a+bx$ over ${\mathbb{F}}_2$ vanishes at $D$, whose residue class of $x$ has irreducible quadratic minimal polynomial. Thus no function has principal divisor $D-2p_{\infty }$, proving its class is nonzero. This establishes the rational points need not generate the degree-zero Picard group qualification with a complete obstruction, while the generation proof above supplies the intended algebraically closed-field answer.

Let $K$ be the function field of the integral scheme $X$ and choose a nonzero rational section of a line bundle $s\in {\mathcal{L}}_{\eta }$. Choose an open trivializing cover $U_i$ with generators $e_i$ for the invertible sheaf $\mathcal{L}$. On each nonempty $U_i$, write $s=a_i{e}_i$ with $a_i\in K^{\ast }$. If $e_i=u_{ij}{e}_j$ on an overlap, then $a_j=a_i{u}_{ij}$, so $a_i/a_j=u_{ij}^{-1}$ is a regular unit. Consequently the local rational functions $a_i$ define a Cartier divisor $D$.

The divisor line bundle ${\mathcal{O}}_X(D)$ restricts to $a_i^{-1}{\mathcal{O}}_{U_i}$. Define its local isomorphism from $\mathcal{L}$ by $e_i\mapsto a_i^{-1}$. On an overlap,so the maps glue. HenceThe rational section maps to the rational function $1$, which also fixes the sign convention for $D$. Choosing a different nonzero rational section changes $D$ by a principal Cartier divisor, so its linear equivalence of Cartier divisors class is unchanged. No projectivity or Noetherian hypothesis is required.