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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 line bundle $M$ on $C\times T$, put $d(t)=\mathrm{deg}{M}_t$, and choose an ample line bundle $A$ on $C$. Fix $t_0$. For sufficiently large $q$, both $H^1(C,M_{t_0}\otimes A^q)$ and $H^1(C,M_{t_0}^{-1}\otimes A^q)$ vanish, by Serre duality and negativity of the degrees of their dual canonical twists. The Semicontinuity theorem for coherent cohomology makes both groups vanish on a neighborhood of $t_0$. The family is flat over $T$: $C\times T\to T$ is flat and the sheaves in question are locally free.

On that neighborhood the Riemann-Roch theorem givesUpper semicontinuity of both zeroth sheaf cohomology dimensions, after shrinking again, gives $d(t)\le d(t_0)$ and $-d(t)\le -d(t_0)$. Hence $d(t)=d(t_0)$. The degree is locally constant, and thus constant on connected $T$. With the usual convention that a variety is irreducible, this proves the assertion as printed. If disconnected varieties are allowed, the necessary qualification is constancy on each connected component: take $T$ to be two points and the bundles ${\mathcal{O}}_{{\mathbb{P}}^1}$ and ${\mathcal{O}}_{{\mathbb{P}}^1}(1)$ to see that a global constant degree need not exist.

Let $C$ be the projective cubic of a Weierstrass equation of an elliptic curve, allowing its elliptic-curve discriminant to vanish, and let $O=[0:1:0]$. Use only the smooth locus of a variety $C_{\mathrm{sm}}(K)$: the singular point, if present, is excluded. For $P,Q$, intersect their chord with $C$, using the tangent if $P=Q$ and counting intersection multiplicity. If the third intersection is $R$, define $P+Q=-R$, wherein general Weierstrass equation of an elliptic curve coordinates. The line through $R$ and $O$ gives this reflection. A vertical chord gives $P+(-P)=O$, and the tangent at $O$ meets $C$ three times at $O$, so $O$ is the identity. The construction is symmetric in $P,Q$. It stays in the nonsingular locus: a line through a singular point has intersection multiplicity at least two there, and therefore cannot also contain two smooth intersections counted with multiplicity.

For the nonsingular case, prove associativity by transporting a known abelian group law. The Abel-Jacobi map of a genus-one curveis bijective over an algebraic closure. Indeed, the Riemann-Roch theorem in genus one says that every divisor class of degree one has a unique effective representative consisting of one point: existence follows from $\ell (D)=1$, and uniqueness follows because two distinct representatives would produce a degree-one map to the projective line, impossible for a genus one curve. Subtracting $O$ gives the claimed bijection.

Any line section represents the same divisor class as $3O$, including tangencies. Thus $P+Q+R\sim 3O$ as divisors on an algebraic curve, while the vertical line gives $R+(-R)+O\sim 3O$. HenceAddition in the Picard group is associative, soThe chord-and-tangent group law is defined over $K$, so the group operation restricts to the $K$-rational points.

For completeness, a singular Weierstrass equation of an elliptic curve produces the smooth-locus group of a singular Weierstrass cubic, not an elliptic curve. Over an algebraic closure, its normalization of an algebraic curve is the projective line; deleting the two preimages of a nodal crossing gives the multiplicative algebraic group, while deleting the single preimage of a cusp gives the additive group. For example, on $y^2=x^3$, the coordinate $u=x/y$, with $u(O)=0$, makes the smooth-locus law addition. On $y^2=x^2(x+1)$ in characteristic different from two, put $t=y/x$ and $z=(t+1)/(t-1)$, with $z(O)=1$; the chord relation gives $z(P)z(Q)z(R)=1$, so the group law is multiplication of $z$. A nonsplit node gives the corresponding form of the multiplicative algebraic group over $K$. These descriptions also establish the singular-case group laws.

The effective group is torsion free and every modular cusp is regular in the preceding sense, so orders of a meromorphic weight-$k$ form are integers. At interior points use a local automorphy trivialization; at a modular cusp use the Fourier order of the appropriate slash transform in its cusp width coordinate. Let $C$ be the sum of all modular cusp points, each once, and setA meromorphic function $\phi$ belongs to the Riemann-Roch space $\mathcal{L}(D(f))$ exactly when $\mathrm{div}\phi +D(f)\ge 0$. In the interior this requires $f\phi$ to have no pole; at a modular cusp it requires order at least one. Conversely, the quotient of any weight-$k$ cusp form by $f$ is a meromorphic weight-zero function satisfying precisely those inequalities. This proves the cusp-form divisor presentationTo compute the degree without imposing a valence formula as an extra assumption, use the meromorphic tensor differential $f^{12}(d\tau {)}^{6k}$. Its automorphy factors cancel. Its order at an interior point is $12{\mathrm{ord}}_{P}f$; at a modular cusp it is $12{\mathrm{ord}}_{P}f-6k$, since $d\tau$ is a nonzero constant times $d{q}_c/q_c$. A meromorphic section of the $6k$th tensor power of the canonical bundle has total divisor degree $6k(2g-2)$. The regular-cusp valence formula on a torsion-free modular curve is thereforeFor $k\ge 3$, $\mathrm{deg}D(f)-(2g-2)=(k-2)d/12>0$. The Riemann-Roch theorem says $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$, and a divisor of negative degree has no nonzero sections. Thus $\ell (K-D)=0$ andUsing $d=(p^2-1)/2$ and $r_{\infty }=p-1$ givesThe canonical-degree and Riemann-Roch facts used here are general results for compact Riemann surfaces, as permitted.

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.

First use the canonical map to obtain the embedding. Its canonical divisor has degree six and $h^0(K_C)=4$. Over an algebraically closed ground field, every effective degree-two divisor on an algebraic curve $E$ satisfies $h^0(E)=1$: a second section would give a nonconstant rational function with poles bounded by $E$, hence a map $C\to {\mathbb{P}}^1$ of degree at most two. Degree one would make $C$ rational, and degree two would make it hyperelliptic, both excluded here.

The Riemann-Roch theorem now givesThus the complete linear system of divisors $|K_C|$ separates every pair of points and every tangent direction, including the tests $E=2p$. By the length-two criterion for a very ample linear system, it is a very ample linear system. Consequently the canonical map embeds $C$ in ${\mathbb{P}}^3$, with degree of a projective curve six. Its image is a nondegenerate projective variety, since the four canonical sections are linearly independent.

The space of homogeneous quadrics in four variables has dimension ten, while the Riemann-Roch theorem givesRestriction therefore has a nonzero kernel, giving a quadric $Q$ containing $C$. This quadric is irreducible: a reducible quadric is a union of two planes, or a double plane, and an integral curve lying in it would lie in a plane, contradicting nondegeneracy.

Similarly, homogeneous cubics form a twenty-dimensional space, whereasThere are at least five independent cubics vanishing on $C$. The multiples of $Q$ by linear forms give only four dimensions, so choose a cubic $F$ vanishing on $C$ which is not divisible by $Q$.

The polynomials $Q,F$ are a regular sequence, hence their intersection $D$ is a projective complete intersection of pure dimension one and degree $2\cdot 3=6$. It contains $C$, which already has degree six. The unmixedness of a complete intersection excludes embedded components. Equal degrees force $C$ to be the only irreducible component and force multiplicity one at its generic point. By generic reducedness with no embedded components, this makes $D$ reduced, so $D=C$ as schemes. We have provedThis is a canonical genus-four curve as a quadric-cubic intersection. The quadric may be smooth or a cone; the argument does not require it to be smooth. The same construction descends over the ground field when the curve is geometrically nonhyperelliptic, since the restriction maps and canonical embedding are defined there.

Let $\mathcal{L}={\mathcal{O}}_C(1)$. Its degree is five. Because the linear span of $C$ has dimension at least three, restrictions of linear forms supply at least four independent global sections, soIf $h^1(C,\mathcal{L})=0$, the Riemann-Roch theorem gives $h^0(\mathcal{L})=6-g$, and therefore $g\le 2$ immediately.

If instead $\mathcal{L}$ is special, the Clifford inequality for curves givesso its integer dimension is at most three, a contradiction. Here is a short proof of the required Clifford inequality for curves, so the bound need not be assumed. Put $r=h^0(\mathcal{L})>0$ and $s=h^0({\omega }_C\otimes {\mathcal{L}}^{-1})>0$. At a smooth geometric point, choose bases of these two section spaces with strictly increasing orders of vanishing $a_1<\cdots <a_r$ and $b_1<\cdots <b_s$. Such bases exist because each successive order-of-vanishing quotient has dimension at most one. The products of the first basis with the first element of the second basis, followed by products of the last element of the first basis with the remaining elements of the second, have ordersThey are linearly independent global sections of ${\omega }_C$, proving the product dimension bound for sections on a curve. Thus $r+s-1\le h^0({\omega }_C)=g$. Combining this with Riemann-Roch theorem, $r-s=\mathrm{deg}\mathcal{L}+1-g$, gives $2r\le \mathrm{deg}\mathcal{L}+2$, which is exactly the displayed Clifford inequality for curves.

The special case is impossible, and henceThis is the genus bound for a nonplanar degree-five curve. The geometric argument can be carried out after extension to an algebraic closure, which preserves these section dimensions.

Here $\mathrm{deg}{K}_C=2$. For $D$, $D-P$ and $D-P-Q$, including $Q=P$, the degrees are $5,4,3$, all greater than two. Thus the correction term in the Riemann-Roch theorem vanishes andThe first drop supplies a section nonzero at every chosen point, proving base-point-freeness of a linear system. For distinct $P,Q$, the second drop supplies a section vanishing at one but not the other, proving point separation. For $Q=P$, it supplies a section vanishing to order exactly one, proving tangent separation. Hence the complete linear system of divisors is very ample and $\boxed{C\hookrightarrow {\mathbb{P}}^3}$ is the induced closed immersion.

The Riemann-Roch theorem states, for a divisor $D$ on a smooth projective curve of genus $g$,where $\ell (D)=\dim H^0(C,\mathcal{O}(D))$ and $K_C$ is a canonical divisor, of degree $2g-2$. Negative-degree divisors have no nonzero global sections. The theorem's usual formulation here is over an algebraically closed field, as in the curve course.

Put $O=0_E$. Choose rational projective coordinates in which $O=(0:1:0)$ and its tangent is $Z=0$. Because $O$ is an inflection point of a plane cubic, that tangent intersects the cubic in the divisor $3O$. Consequently its cubic equation has the formHere $a\ne 0$, since the restriction to $Z=0$ is a nonzero multiple of $X^3$, and $b\ne 0$, since smoothness at $O$ requires a nonzero $Z$-derivative there. Completing the square in $Y$, by the rational linear change $Y^{\prime }=Y+(cX+eZ)/(2b)$, gives on $Z=1$Set $x^{\prime }=c_{0}x$ and $y^{\prime \prime }=c_0{y}^{\prime }$. ThenThese are invertible rational projective coordinate changes, and $O$ remains the unique point at infinity. Smoothness ensures that the resulting monic cubic has no repeated root. Thus this is a Weierstrass equation of an elliptic curve, without requiring any extraction of a square or cube root in $\mathbb{Q}$.

On this model write $-(x,y)=(x,-y)$ and $-O=O$. To add $P,Q$, take their joining line, or the tangent if $P=Q$, and let $R$ be the third intersection counted with multiplicity. DefineFor a vertical line its third intersection is $O$, so $P+(-P)=O$. The line through $O$ and an affine point is vertical, giving $P+O=P$. The tangent at $O$ has triple contact, giving $O+O=O$. This is the chord-and-tangent group law.

If $P,Q$ are rational, their joining line or tangent is rational. Its residual intersection point is rational: substituting the line in the cubic leaves a cubic whose other two roots, counted with multiplicity, are rational. Thus addition preserves $E(\mathbb{Q})$. The construction is symmetric in $P,Q$, so it is commutative.

For associativity, use divisor classes. A smooth plane cubic has geometric genus $1$, by the genus-degree formula. The Riemann-Roch theorem says that a divisor of degree $1$ on a genus-one curve has a one-dimensional space of sections. Therefore every degree-zero divisor class is uniquely represented by $P-O$: add $O$, take its unique effective degree-one representative, and then subtract $O$. Uniqueness also follows because a nonconstant function with its only pole a simple pole at $Q$ would contradict $\ell (Q)=1$.

Every line section is linearly equivalent to the tangent section $3O$. Thus, whenever $P,Q,R$ are collinear, including tangencies,A vertical section gives $[(-R)-O]=-[R-O]$. Hence the chord-and-tangent group law becomes ordinary addition in the degree-zero Picard group under the injective correspondence $P\mapsto [P-O]$. Associativity follows from associativity in that abelian group. Together with closure, identity and inverses already checked, this proves that $E(\mathbb{Q})$ is an abelian group.

For a nonsingular projective integral curve, apply the Riemann-Roch theorem to the divisor on an algebraic curve $mP$. It gives $\ell (mP)\ge m\mathrm{deg}P+1-g$, where $g$ is the geometric genus. For sufficiently large $m$, this exceeds the dimension of the space of constant functions; over the usual geometrically integral convention that dimension is one. Thus $L(mP)$ contains a nonconstant rational function.

By the definition of $L(mP)$, the principal divisor on an algebraic curve of that function plus $mP$ is effective, so it has no pole away from $P$. Therefore there is a nonconstant rational function regular on $X\setminus \{P\}$. If $P$ is not a rational point, using its closed-point degree in the same inequality handles that case. A positive pole order at $P$ is unavoidable, since regular functions on a projective integral curve are constant in its constant field.

Because $|D|$ has no fixed components, choose two members $G_1,G_2$ with no common component. This uses the infinitude of the algebraically closed field and avoidance of finitely many proper linear subspaces of the section space. On a smooth algebraic surface, their intersection number is the sum of local intersection multiplicities. Since $G_1\cdot G_2=D^2=0$, they are disjoint. A basepoint of $|D|$ would belong to both, so $|D|$ is basepoint-free.

The resulting morphism has nonconstant image since $\dim |D|\ge 1$. Its image cannot be a surface: a generically finite morphism defined by $D$ would make $D^2$ the positive product of its degree and the degree of its image. Thus the image is a curve. Apply Stein factorization to obtain $g:X\to C$ with connected fibers and $g_{\ast }{\mathcal{O}}_X={\mathcal{O}}_C$, where $C$ is a smooth projective curve. There is a positive-degree line bundle $L$ on $C$ with ${\mathcal{O}}_X(D)=g^{\ast }L$. The projection formula for sheaves and Riemann-Roch theorem give $h^0(X,mD)=h^0(C,L^m)=m\mathrm{deg}L+O(1)$. Therefore

For a divisor on an algebraic curve $D$ on a smooth projective curve $C$ of geometric genus $g$, and a canonical divisor $K_C$, the Riemann-Roch theorem states

Choose a point $Q\in E$. Since $g(E)=1$, a nonzero regular differential has an effective divisor of degree $2g-2=0$, so $K_E\sim 0$. The Riemann-Roch theorem therefore givesChooseThe rational functions $x$ and $y$ have their only poles at $Q$, of orders two and three respectively.

The seven functionsbelong to the six-dimensional Riemann-Roch space $L(6Q)$, so they satisfy a nontrivial linear relation. Comparing pole orders shows that the coefficients of both $x^3$ and $y^2$ are nonzero. After rescaling, the relation has the Weierstrass formThe divisor $3Q$ is very ample, so its complete linear system embeds $E$ in ${\mathbb{P}}^2$ and the displayed relation cuts out its image; this is the Weierstrass construction from Riemann-Roch in genus one.

Because $\mathrm{char}k\ne 2$, completing the square removes the terms linear in $y$, and because $\mathrm{char}k\ne 3$, translating $x$ removes the quadratic term. Thus the equation becomesSince $k$ is algebraically closed, the cubic factors asSmoothness of $E$ says that the elliptic-curve discriminant is nonzero, equivalently the three roots are distinct. Hence

DefineIf $D$ is a divisor class of degree zero, Riemann--Roch gives $\ell (D+Q)=1$, because $K_E\sim 0$ and the divisor $-D-Q$ has negative degree. A nonzero section therefore yieldsfor some point $P$, proving that $\alpha$ is surjective. If $\alpha (P)=\alpha (P^{\prime })$, then $P-P^{\prime }$ is a principal divisor. For $P\ne P^{\prime }$ its defining function would have one simple pole and give a degree-one map $E\to {\mathbb{P}}^1$, impossible for a genus-one curve. Thus $P=P^{\prime }$. The map and its inverse are regular, so this proves the Abel-Jacobi map of a genus-one curve is an isomorphism.

Transport the addition of the divisor class group through $\alpha$: define $P⊞R$ byThen $Q$ is the identity, and this is the chord-and-tangent group law on the plane cubic.

For the displayed equation, inversion is $(x,y)\mapsto (x,-y)$. Thus $P⊞P=Q$ exactly when $P=-P$. Besides the point at infinity $Q$, this requires $y=0$. Therefore the complete set of two-torsion points is

For $N>2g-2$, the divisor $K_X-Np$ has negative degree, so $\ell (K_X-Np)=0$. The Riemann-Roch theorem givesStarting from $\ell (0)=1$ and using part (ii), each integer $1\le n\le N$ contributes one to $\ell (Np)$ exactly when it is not a gap. ThereforeComparison with the preceding formula gives $|G(p)\cap \{1,\dots ,N\}|=g$. Moreover every $n>N$ is a nongap because the Riemann--Roch formula makes every subsequent dimension jump equal to one. Thus the Weierstrass gap theorem says