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For vector bundles on a smooth projective curve, a dualizing sheaf is a line bundle $D_C$ with a trace $H^1(C,D_C)\to k$ making the evaluation pairings $H^i(C,E)\times H^{1-i}(C,E^{\vee }\otimes D_C)\to k$ perfect for $i=0,1$. Choose a finite generically separable map to the projective line. The trace-dual module for this finite map, tensored with ${\Omega }_{{\mathbb{P}}^1}^1$, is ${\Omega }_C^1$ by the different–differential identity. Finite pushforward and residue duality on the projective line then prove the pairings, rather than assuming the curve-duality theorem to establish them.

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.

Cover the projective line by $U_0=\mathrm{Spec}k[t]$ and $U_1=\mathrm{Spec}k[t^{-1}]$. On their overlap the coordinate is invertible. Choose frames $e_0,e_1$ of the twisting sheaf on projective space $\mathcal{O}(n)$ with $e_1=t^n{e}_0$. The vanishing of quasi-coherent cohomology on an affine scheme applies to both charts and their intersection, so the Čech cochain complexcomputes the sheaf cohomology. Its kernel identifies with $k[t]\cap t^{n}k[t^{-1}]$, and its cokernel isFor $n\ge 0$, the intersection has basis $1,t,\dots ,t^n$; for $n<0$ it is zero. In the quotient, powers $t^j$ with $j\ge 0$ or $j\le n$ disappear. The surviving basis for $n\le -2$ is $t^{n+1},\dots ,t^{-1}$; for $n\ge -1$ none survives. HenceThis is the Čech cohomology of twists on the projective line.

By the preceding question, $\omega ={\Omega }_{{\mathbb{P}}^1/k}^1=\mathcal{O}(-2)$. In the $dt$ frame its degree-one Čech cohomology is one-dimensional, represented by $t^{-1}dt$. DefineThis is well defined: differentials regular on $U_0$ have nonnegative powers, and those regular on $U_1$ have powers at most $-2$ when expressed using $dt$. Evaluation and the cup product give natural pairingsFor $E=\mathcal{O}(n)$ with $n\ge 0$, the basis element $t^a$ in $H^0(E)$ pairs with $t^{-a-1}dt$ in $H^1(E^{\vee }\otimes \omega )$ and with no other element of this dual Laurent basis. The pairing matrix is a permutation matrix, hence invertible. For $n\le -2$ the same computation exchanges degrees zero and one. For $n=-1$ both sides vanish. Thus the pairings are perfect for every line bundle on ${\mathbb{P}}^1$.

Here we used the classification of line bundles on the projective line: they are $\mathcal{O}(a)$ for integers $a$. One direct justification is that every line bundle has a nonzero rational section and hence is associated with a divisor on an algebraic curve. The finite closed point defined by an irreducible polynomial $p(t)$ is linearly equivalent to $(\mathrm{deg}p)\infty$, using the principal divisor of $p(t)$. Thus every divisor is equivalent to its degree times $\infty$, proving the classification over arbitrary $k$.

To extend the pairing to every locally free coherent sheaf $E$, induct on rank, without assuming the splitting theorem of the next question. A one-dimensional subspace of the generic fibre, saturated in $E$, gives an exact sequencewith $L$ a line bundle and $Q$ a lower-rank vector bundle. This saturated line subbundle on a smooth curve exists because its stalk and its torsion-free quotient are free over the discrete valuation rings of the curve. The dual sequence, tensored with $\omega$, is exact as well, since the original sequence splits locally.

The two long exact sequences in sheaf cohomology, after dualizing one, are compatible with the evaluation pairings. In Čech cohomology, this compatibility follows by lifting a section locally: the differences of its lifts represent its connecting class, and contracting those differences with a dual cocycle gives the transpose connecting map, with the usual sign. Thus the connecting maps are adjoint; changing one vertical map's sign gives a commuting diagram. Its rows areThe maps for $L$ and $Q$ are isomorphisms by the line-bundle calculation and induction. Exactness then makes the maps for $E$ isomorphisms, by the Five lemma (or by comparing the relevant kernels and cokernels). All groups are finite-dimensional because the curve is projective and the sheaves coherent. Therefore residue duality on the projective line provesThese isomorphisms are natural in $E$. Higher sheaf cohomology vanishes by the same two-chart complex.

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.