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For a smooth plane quartic $C\subset {\mathbb{P}}^2$, the adjunction formula gives $K_C\cong {\mathcal{O}}_C(1)$. Hence every line section $C\cap \ell$ is a canonical divisor.

Every nonzero holomorphic differential form on a compact Riemann surface of genus one is nowhere zero. Its canonical divisor is an effective divisor of degree $2g-2=0$, and hence has every coefficient zero. In particular, such a differential trivializes the canonical bundle. This assertion concerns the differential itself, not the individual factors in a representation $fdh$.

A divisor on an algebraic curve is a finite formal sumIts degree of a divisor is $\mathrm{deg}D={\sum }_p{n}_p$. It is an effective divisor, written $D\ge 0$, when every $n_p\ge 0$.

A meromorphic differential on a Riemann surface is a one-form represented in every local holomorphic coordinate $t$ by $a(t)dt$, where $a$ is a meromorphic function. Under a coordinate change $t=t(u)$, its coefficient becomes $a(t(u))t^{\prime }(u)$, so these expressions agree on overlapping charts. Since $t^{\prime }(u)$ is a nonvanishing holomorphic function for a valid coordinate change, the order of a zero or pole of the differential is independent of coordinate. Equivalently, for a fixed nonconstant meromorphic function $h$, any such differential can be written $fdh$ with $f$ meromorphic.

A canonical divisor is the divisor on an algebraic curveof a nonzero meromorphic differential on a Riemann surface $\eta$. If ${\eta }^{\prime }=f\eta$, then $({\eta }^{\prime })=(\eta )+(f)$, so all canonical divisors are related by linear equivalence of divisors. Their degree is

The Riemann-Roch theorem states that, for every divisor on an algebraic curve $D$ and any canonical divisor $K$ on the nonsingular projective curve $C$ of genus $g$,The term $\ell (K-D)$ measures the failure of the naive count of conditions on meromorphic functions.

The generator $\eta =fdh$ is a nonzero holomorphic differential form. Its canonical divisor $(\eta )$ is an effective divisor, since there are no poles. In genus one, the degree of a divisor formula in part (a) givesAn effective divisor of degree zero has every coefficient zero. ThusThis proves nonvanishing of holomorphic differentials in genus one. It concerns the differential itself: a zero or pole of the particular coefficient $f$ may be cancelled by the local behavior of $dh$.

Apply Riemann-Roch at $D=K$ and use $\ell (K)=g$ and $\ell (0)=1$:Rearranging gives $\mathrm{deg}K=2g-2$. This holds for every canonical divisor, since any two differ by a principal divisor of degree zero.

A divisor on an algebraic curve is a finite integer sum $D={\sum }_p{n}_{p}p$; its degree is ${\sum }_p{n}_p$. For a meromorphic function its principal divisor records its zero and pole orders, andLet $K$ be the divisor of a nonzero meromorphic differential, called a canonical divisor. The Riemann-Roch theorem statesMultiplication by that differential identifies $L(K)$ with the space of holomorphic differentials. By the preceding result $L(0)$ consists exactly of constants, so $\ell (0)=1$. Applying the formula at $D=0$ gives $1-\ell (K)=1-g$, hence $\ell (K)=g$.

A divisor on an algebraic curve is a finite formal sum $D=\sum n_{p}p$ with integer coefficients. Its degree is $\mathrm{deg}D=\sum n_p$. The principal divisor of a nonzero meromorphic function is $(f)=\sum {\mathrm{ord}}_p(f)p$. For a nonconstant $f$, zeros and poles are fibers over zero and infinity of the same map to the sphere, each of total multiplicity equal to its degree; hence $\mathrm{deg}(f)=0$. A nonzero constant has zero divisor.

Write $L(D)=\{f:(f)+D\ge 0\}\cup \{0\}$ and $\ell (D)=\dim L(D)$. The Riemann-Roch theorem stateswhere $K_C$ is a canonical divisor. Taking $D=p_1+\cdots +p_{g+1}$ gives $\ell (D)=2+\ell (K_C-D)\ge 2$. Constants span only one dimension, so there is a nonconstant $f\in L(D)$. It has no poles outside these points and at most a simple pole at each. Thus the required nonconstant function exists, with actual poles at some nonempty subset because a holomorphic function on a compact connected curve is constant.

A divisor on an algebraic curve is a finite formal integer sum $D={\sum }_p{n}_{p}p$. Its degree is ${\sum }_p{n}_p$. The Riemann-Roch space $L(D)$ consists of zero and the meromorphic functions $h$ with $(h)+D\ge 0$, and $\ell (D)=\dim L(D)$. The Riemann-Roch theorem stateswhere $K$ is a canonical divisor, of degree $2g-2$. In genus one, $\mathrm{deg}(K-D)=-n<0$ for $\mathrm{deg}D=n>0$. A negative-degree divisor on an algebraic curve has no nonzero section, since a principal divisor has degree zero and an effective divisor has nonnegative degree. HenceA nonsingular plane cubic has genus one by the genus-degree formula. Therefore $\ell (p+q)=2$, and there is a nonconstant $f\in L(p+q)$. It has at most simple poles at $p,q$ and no other poles. It cannot have no poles, since a holomorphic function on a compact Riemann surface is constant. It cannot have just one simple pole either, by the allowed fact, or by $\ell (p)=\ell (q)=1$. Thus $f$ has simple poles at both $p$ and $q$, and no others.

The genus-degree formula for a nonsingular plane curve givesA divisor is a finite formal sum $D={\sum }_p{n}_p[p]$ with integer coefficients; its degree is $\mathrm{deg}D={\sum }_p{n}_p$. An effective divisor has all $n_p\ge 0$. A hyperplane divisor is the intersection with a projective line not containing the curve, counted with local intersection multiplicities. Its degree is $d$.

Define $L(D)=\{0\}\cup \{f\text{ meromorphic}:(f)+D\ge 0\}$ and $\ell (D)={\dim }_{\mathbb{C}}L(D)$. For a canonical divisor $K$, the Riemann-Roch theorem statesHere $(f)$ records zero orders positively and pole orders negatively, and $\mathrm{deg}K=2g-2$.

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.

Let $g$ be the geometric genus of $X$. Choose an integer $N\ge 2g+1$. For each prescribed point $x_i$, both $N{x}_i$ and $(N-1)x_i$ have degree greater than $2g-2$, so their complementary canonical divisors have negative degree and their complementary Riemann-Roch spaces vanish. The Riemann-Roch theorem therefore givesChoose $f_i\in L(N{x}_i)\setminus L((N-1)x_i)$. This is possible because the two dimensions differ by one. Such a rational function has its only pole at $x_i$, and its pole order there is exactly $N$. Set $f={\sum }_{i=1}^m{f}_i$.

At $x_i$ all terms except $f_i$ are regular. In a local parameter, the term of order $-N$ in $f_i$ has a nonzero coefficient, and none of the other terms has a negative-power term. Hence that coefficient cannot cancel. Away from the prescribed points every summand is regular. We obtain the stronger conclusionThus $f$ is regular on $U$ and has a pole at every $x_i$. This rational functions with prescribed poles construction works in every characteristic and does not require a generic-coefficient choice. If the prescribed list is empty, a constant function meets the vacuous requirements.

Let $E$ be the elliptic curve, and choose a point $O\in E(k)$ to serve as identity. Its geometric genus is one and its canonical divisor $K$ has degree zero. For any divisor $D$ of degree one, the Riemann-Roch theorem givesSince $K-D$ has degree $-1$, its Riemann-Roch space is zero: a nonzero section would produce an effective divisor of negative degree. Hence $\ell (D)=1$. Choose a nonzero $f\in L(D)$. The divisor $D+(f)$ is effective of degree one, so over the algebraically closed field it is a single point $P$. Thus every degree-one divisor class has a point representative.

This representative is unique. If $P$ and $Q$ are linearly equivalent, choose $h$ with $(h)=P-Q$. Then $h\in L(Q)$. That space has dimension one and already contains the constants, so $h$ is constant and $P=Q$. Consequentlyis injective. It is also surjective: if $D$ represents a degree-zero divisor class, apply the preceding result to $D+O$ to obtain its unique point representative $P$, giving $[D]=[P-O]$.

The degree-zero Picard group of a curve is a group because divisors add commutatively, principal divisors form a subgroup, and degree passes to the quotient. DefineEquivalently, $R=P\oplus Q$ is the unique point satisfying the linear equivalence of divisors $P+Q\sim R+O$, where the plus signs in this equivalence denote divisor addition. The zero class corresponds to $O$, so $O$ is the identity. The inverse of $P$ is the unique point $P^{\prime }$ with $P+P^{\prime }\sim 2O$. Associativity follows because both $(P\oplus Q)\oplus R$ and $P\oplus (Q\oplus R)$ map to $\iota (P)+\iota (Q)+\iota (R)$, and $\iota$ is injective; commutativity follows in the same way. Thus this is an abelian group structure on $E(k)$, obtained directly by the elliptic curve group law from Riemann-Roch. The identity choice is part of the construction.

A divisor on a compact connected Riemann surface is a finite formal sum $D={\sum }_p{n}_{p}p$ with integer coefficients; its degree is ${\sum }_p{n}_p$. A canonical divisor is the divisor of zeros and poles of a nonzero meromorphic differential. Two divisors are linearly equivalent when their difference is the divisor of a nonzero meromorphic function. Such principal divisors have degree zero, but equal degree does not generally imply linear equivalence of divisors.

For example take distinct points $p,q$ on a genus-one surface. If $p-q$ were principal, its function would define a degree-one meromorphic map to the sphere, since it has just one simple pole. A degree-one map between compact Riemann surfaces is a biholomorphism, contradicting their different genera. Thus these degree-one divisors are not equivalent.

Define $L(D)=\{f\text{ meromorphic}:(f)+D\ge 0\}\cup \{0\}$ and $\ell (D)={\dim }_{\mathbb{C}}L(D)$. The Riemann-Roch theorem statesSince all holomorphic functions on the compact surface are constant, $\ell (0)=1$. Setting $D=0$ gives $\ell (K)=g$. Multiplication by a fixed meromorphic differential identifies $L(K)$ with holomorphic differentials, so their dimension is $g$. Setting $D=K$ then givesFor $g=2$, choose a nonzero holomorphic differential, whose effective canonical divisor has degree two. Since $\ell (K)=2$, $L(K)$ contains a nonconstant meromorphic function. Its pole divisor is bounded by that effective $K$, proving existence of a nonconstant function with total pole multiplicity at most two.

Choose a Weierstrass equation of an elliptic curve, with origin $O$ at infinity. For $P,Q$, draw their chord, or the tangent if they coincide, and call its third intersection $R$, counted with intersection multiplicity. The line through $R$ and $O$ gives a third point $S$; define $P+Q=S$. This is the chord-and-tangent group law. The inverse of an affine point on $y^2+a_{1}xy+a_{3}y=x^3+a_2{x}^2+a_{4}x+a_6$ is $(x,-y-a_{1}x-a_3)$, obtained from its vertical line through $O$. These conventions give identity $O$, and cover tangent and flex cases by multiplicity.

Here is a proof of associativity, without assuming that the geometric construction already defines a group. Work first over an algebraic closure and use divisors on an algebraic curve. The Riemann-Roch theorem says $\ell (D)-\ell (K_E-D)=\mathrm{deg}D+1-g$. For a smooth genus-one curve the canonical divisor has degree zero, so every degree-one divisor has $\ell (D)=1$. Thus every divisor class of degree one has a unique effective representative, necessarily a single point. Existence follows from a nonzero section; uniqueness follows because two distinct effective representatives would give linearly independent sections in the same one-dimensional space. Consequentlyis a bijection: add $(O)$ to any degree-zero divisor class, and use its unique degree-one representative.

A line through $P,Q,R$ cuts out a divisor linearly equivalent to $3(O)$, since the line at infinity has triple intersection at the Weierstrass flex $O$. Equivalently, the ratio of the line equation to that of the line at infinity has principal divisor $(P)+(Q)+(R)-3(O)$. Thus the three corresponding degree-zero classes sum to zero. Applying the same fact to $O,R,S$ shows that the point produced by the chord-and-tangent process satisfiesAddition of divisors modulo principal divisors is associative. The two parenthesizations of $P+Q+T$ therefore have identical classes; injectivity of the displayed bijection makes them the same point. Hence the chord-and-tangent law is associative. This is associativity of the chord-and-tangent law via divisor classes. Its construction is defined over the original field, so the same conclusion holds for its rational points.

For a smooth projective curve of genus $g$, the Riemann-Roch theorem stateswhere $D$ is a divisor, $K$ a canonical divisor, and $\ell (D)=\dim H^0(V,{\mathcal{O}}_V(D))$. In particular, applying it to $D=0$ and $D=K$ gives $\ell (K)=g$ and $\mathrm{deg}K=2g-2$.

For a nonconstant morphism $f:C\to D$ of degree $d$ in characteristic zero, choose local parameters with $t\circ f=u^e$ up to a unit at a point of ramification index $e$. Pulling back a differential multiplies its local order by $e$ and adds $e-1$, because $d(u^e)=e{u}^{e-1}du$ and $e\ne 0$ in the field. ThusThe degree of a divisor of a pulled-back divisor is $d$ times its degree. Taking degrees and using the Riemann-Roch theorem gives the Riemann-Hurwitz formula

A smooth plane cubic has genus one by the genus-degree formula. In the displayed normal form, $F(0,X_1)=c{X}_1^3$ with $c\ne 0$: otherwise $X_0$ divides $F$ and the cubic is reducible. Therefore the line $X_0=0$ meets the cubic in the divisor $3P_0$, and ${\mathcal{O}}_V(1)\cong {\mathcal{O}}_V(3P_0)$. The elliptic curve group law identifies $P$ with the degree-zero divisor class $[P-P_0]$. Order three means $3P\sim 3P_0$. Consequently ${\mathcal{O}}_V(3P)\cong {\mathcal{O}}_V(1)$ and has a section with zero divisor $3P$.

Every section of ${\mathcal{O}}_V(1)$ is the restriction of a linear form: ambient linear forms form a three-dimensional space and restrict injectively, while Riemann-Roch theorem gives $h^0({\mathcal{O}}_V(1))=3$, since $\mathrm{deg}(K-3P_0)=-3$ forces the dual term to vanish. Thus this section is some nonzero $H$, proving $V\cap V(H)=\{P\}$, with intersection multiplicity three. This is the three-torsion points are flexes of a plane cubic property.

Projection forgetting $X_3$ defines a morphism $W\to V$: its only possible indeterminacy point $(0:0:0:1)$ does not satisfy the second equation. It is a double cover. Away from the zeros of $H_1{H}_2$, its two square roots are distinct. Each line intersects $V$ in three simple points, since it is nowhere tangent; the two triples are disjoint because the lines do not meet on $V$. In a local trivialization near any of these six points the covering equation is $s^2=ua(u)$ with $a(0)\ne 0$, so the ramification index is two. There is no other ramification. The Riemann-Hurwitz formula therefore gives

For a divisor $D$ on a smooth projective curve, let $L(D)=\{f\in k(X):\mathrm{div}(f)+D\ge 0\}\cup \{0\}$, and write $\ell (D)={\dim }_{k}L(D)$. The Riemann-Roch theorem stateswhere $g$ is the genus, $K_X$ is a canonical divisor given by a nonzero rational differential, and the degree of a divisor is the sum of its coefficients since $k$ is algebraically closed.

For genus one, a nonzero regular differential has a nonnegative divisor of degree $2g-2=0$, hence no zeros; therefore $K_X\sim 0$. Negative-degree divisors have no nonzero sections. Thus $\ell (n{P}_{\infty })=n$ for $n>0$, $\ell (0)=1$, and $L(n{P}_{\infty })=0$ for $n<0$.

Choose $x\in L(2P_{\infty })\setminus L(P_{\infty })$ and $y\in L(3P_{\infty })\setminus L(2P_{\infty })$. Their unique poles have orders two and three respectively. Distinct pole orders prove linear independence in the following complete list:Since $y^2$ also lies in the last six-dimensional space, it satisfiesThe coefficient $a$ cannot vanish because only $x^3$ among the displayed basis has a sixth-order pole available to cancel the leading pole of $y^2$. Homogenizing gives a plane cubic relation, valid also in characteristics two and three without completing squares or dividing by three.

It remains to prove that the map actually embeds the curve. For $D=3P_{\infty }$, Riemann-Roch gives $\ell (D)=3$, $\ell (D-Q)=2$ for every point $Q$, and $\ell (D-Q-R)=1$, including $R=Q$. The first equality makes the linear system base-point-free; the second separates distinct points, and the double-point case separates tangent directions. Therefore $D$ is very ample and ${\varphi }_D$ is a closed immersion into ${\mathbb{P}}^2$. Its pullback of a line has degree three, so its image has degree three. The displayed cubic equation consequently defines the image, which is smooth because it is isomorphic to $X$. Thus $\boxed{{\varphi }_{3P_{\infty }}:X\cong \text{a smooth plane cubic}}$. The affine expression $(1:x:y)$ extends across $P_{\infty }$ to $(0:0:1)$ by the pole orders.

An elliptic curve works over every algebraically closed field, including one algebraic over a finite field. Give it explicitly in ${\mathbb{P}}^2$ byIn the first case the only point at infinity is $O=[0:1:0]$, where the derivative with respect to $z$ is nonzero. In the affine chart $z=1$, a singular point of an algebraic variety would have $y=0$, $x^3-x=0$, and $3x^2-1=0$. The possible roots $x=0,1,-1$ cannot satisfy the last equation in characteristic different from $2$: at $0$ it is $-1$, and at $\pm 1$ it is $2$. In the second case the three homogeneous partial derivatives are $x^2,z^2,y^2$, up to harmless signs, and cannot vanish simultaneously at a projective point. Thus both cubics are smooth varieties. A reducible plane cubic would have intersecting positive-degree components by Bézout theorem, and be singular at an intersection; therefore these smooth cubics are irreducible varieties. The genus-degree formula gives $g=(3-1)(3-2)/2=1$. With the marked point $O$ they are elliptic curves.

For a smooth algebraic curve, its divisor class group is its Picard group. Its degree-zero subgroup isThe map $P\mapsto [P-O]$ gives an isomorphism of abelian groupsTo recall why it is a bijection, Riemann-Roch theorem in geometric genus one says every degree-one divisor class has exactly one independent section: a canonical divisor has degree zero, so the dual degree-$-1$ term has no section. Thus the class contains one effective divisor on an algebraic curve of degree one, namely a point $P$, and that point is unique. Translating by $-O$ gives the claimed bijection; this is precisely the chord-and-tangent group law on the elliptic curve.

Choose a prime integer $\ell$ different from $\mathrm{char}k$. The multiplication-by-n morphism $[\ell ]:E\to E$ has derivative $\ell$ times the identity at $O$, so it is nonconstant. A nonconstant morphism of algebraic varieties between smooth projective curves is surjective: its image is closed by proper morphism and cannot have algebraic dimension zero. Since $k$ is an algebraically closed field, this makes $[\ell ]$ surjective on $k$-points. ThereforeYet $E(k)$ is infinite. For each $x\in k$, the affine equation has a solution for $y$ in the algebraically closed field; since $k$ is infinite, this gives infinitely many points.

By the Fundamental theorem of finitely generated abelian groups, a finitely generated abelian group $A$ satisfying $A=\ell A$ has no free part and is a finite torsion group of order prime to $\ell$. The infinite group $E(k)$ cannot have this property while being finitely generated. Thus ${\mathrm{Pic}}^0(E)$ is not finitely generated. A subgroup of a finitely generated abelian group is finitely generated, soThis proof uses divisibility and infinitude, rather than uncountability, so it remains valid over countable algebraically closed fields.

For a smooth projective curve $C$, let $g=h^1(C,{\mathcal{O}}_C)$ be its geometric genus, let $K_C$ be a canonical divisor, and write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ for the dimension of a vector space of the Riemann-Roch space. The Riemann-Roch theorem saysfor every divisor on an algebraic curve $D$ on $C$.

Here is the cohomological proof of Riemann-Roch for curves. At a closed point $P$, the local ring is a discrete valuation ring. A local uniformizer shows that increasing the allowed pole order by one gives the short exact sequence of sheavesThe last term is a length-one skyscraper sheaf; its precise identification with $k$ is noncanonical, but its sheaf cohomology has $h^0=1$ and $h^1=0$. All the relevant groups of sheaf cohomology are finite-dimensional because $C$ is a projective variety, and coherent sheaf cohomology above degree one vanishes on a algebraic curve. The long exact sequence in sheaf cohomology therefore givesSince $k$ is an algebraically closed field, every closed point has degree one. Iterating this identity for both positive and negative coefficients of $D$ provesEvery global regular function on a projective variety which is an irreducible variety is constant; hence $h^0(C,{\mathcal{O}}_C)=1$ and $\chi ({\mathcal{O}}_C)=1-g$.

Finally apply Serre duality for a smooth projective curve:Taking dimensions of a vector space and inserting this in the Euler characteristic of a coherent sheaf identity proves the displayed Riemann-Roch theorem. The substantial inputs from sheaf cohomology are finiteness and vanishing for coherent sheaves on a projective curve, and Serre duality; the change of Euler characteristic of a coherent sheaf is proved directly by the point exact sequence. In particular the argument handles negative as well as effective divisors on an algebraic curve.

The Appell–Humbert theorem identifies the positive Hermitian form $H_1$ with a holomorphic line bundle whose First Chern class is represented, in our linear-first convention, byIts integral over the square fundamental domain is $1$, so $\mathrm{deg}L=1$. In particular the oriented Chern pairing is $-E_1(1,i)=1$; using $E_1(1,i)$ without adjusting the Hermitian convention would give the wrong sign.

The complex torus has a nowhere-vanishing differential $dz$, so its canonical divisor is trivial and its genus is one. A nonzero section of a line bundle of negative degree would have an effective divisor of negative degree, which is impossible. Thus $H^0(X,L^{-1})=0$. The Riemann-Roch theorem and Serre duality giveTherefore $\boxed{h^0(X,L)=1}$. The choice of semicharacter of a complex lattice changes the degree-zero component of the line bundle, but not this calculation.

For a smooth projective curve $C$ of geometric genus $g$ and a divisor on an algebraic curve $D$, write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ and let $K_C$ be a canonical divisor. The curve form of the Riemann-Roch theorem isHere $\mathrm{deg}{K}_C=2g-2$ and $\ell (K_C)=g$. In line bundle form,The usual curve convention is a smooth geometrically integral projective curve; geometric embedding tests may be checked after extending the ground field to an algebraic closure.