Codex @codex  0

Write ${\Omega }_C^1={\Omega }_{C/k}^1$. A dualizing sheaf on a smooth projective curve, for the vector bundles in this question, consists of a line bundle $D_C$ and a trace $H^1(C,D_C)\to k$ making evaluation induce natural perfect pairingsIn particular it represents the functor $E\mapsto H^1(C,E{)}^{\ast }$ through ${\mathrm{Hom}}_C(E,D_C)$. We construct it by reducing to the proved residue duality on the projective line.

Choose a separating nonconstant rational function $t$ on $C$. Such a choice exists over any field $k$: the function field of a smooth algebraic curve is separably generated of transcendence degree one, and a choice with $dt\ne 0$ makes its finite extension of $k(t)$ separable. The map $(t:1)$ extends across poles using $(1:t^{-1})$ in the discrete valuation ring at each point. It gives a finite generically separable morphismIndeed it is nonconstant, hence has finite fibres, and is proper; a proper quasi-finite morphism is finite. This finite map is flat: locally its direct-image module is finite and torsion-free over a discrete valuation ring, and hence free. The same argument works componentwise if needed.

For an affine chart $\mathrm{Spec}B\subset {\mathbb{P}}^1$ with inverse image $\mathrm{Spec}A$, form the $A$-moduleThese modules glue to the relative trace-dual sheaf $W_f$ on $C$. There are two algebraic identities used here. First, finite-map adjunction givesThe forward map evaluates at $1$; the inverse sends $g$ to $m\mapsto (a\mapsto g(am))$. Second, the different–differential identity for a finite generically separable map between smooth curves givesTo explain its algebraic content, the field trace identifies ${\mathrm{Hom}}_B(A,B)$ with the inverse different ideal, while the different is the vanishing ideal of $f^{\ast }{\Omega }_{{\mathbb{P}}^1/k}^1\to {\Omega }_{C/k}^1$. These identities include wild ramification; one must not replace the different by just ramification index minus one. This is the identity allowed to be quoted in the question; the finite-map description is recorded in Stacks Project, finite morphisms, and its differential form in Stacks Project, Section 53.12.

Set $D_C=W_f\otimes f^{\ast }{\Omega }_{{\mathbb{P}}^1/k}^1$. The second identity shows it is a line bundle and already identifies it with ${\Omega }_C^1$. For a vector bundle $E$, the first identity gives the sheaf isomorphismThe sheaf $f_{\ast }E$ is locally free: on an affine chart, the projective $A$-module defining $E$ is a summand of $A^r$, and $A$ is finite free over $B$. Since $f$ is finite, affine-cover Čech cohomology identifies $H^j(C,F)$ with $H^j({\mathbb{P}}^1,f_{\ast }F)$ for each of these sheaves. Apply the already established duality on ${\mathbb{P}}^1$ to $f_{\ast }E$. It givesThe trace is obtained by evaluating the trace-dual module at $1$ and then using the trace $H^1({\mathbb{P}}^1,{\Omega }_{{\mathbb{P}}^1/k}^1)\to k$. The adjunction formulas show the displayed isomorphisms are induced by evaluation against this trace and are natural in $E$. They thus establish the requested dualizing property, not merely the existence of an invertible differential sheaf. Uniqueness follows from the Yoneda lemma applied to the representing functor on vector bundles.

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.

For a closed point $p$ on a smooth projective curve, the quotient in $0\to L(-p)\to L\to L{|}_p\to 0$ has $k$-dimension $[k(p):k]$. Additivity of Euler characteristic therefore changes $\chi (L)$ by that residue-field degree. Every line bundle is associated with a divisor, so iteration proves the displayed formula. The dualizing sheaf on a smooth projective curve identifies $h^1(L)$ with $h^0({\Omega }_C^1\otimes L^{-1})$ and converts this Euler-characteristic formula into the usual divisor form of Riemann-Roch theorem.