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Let $E$ be a genus one curve over an algebraically closed field and choose $O\in E$. The mapis an isomorphism. For a degree-zero divisor $D$, Riemann--Roch gives a nonzero section of $D+O$, so $D\sim P-O$ for some point $P$. If $P-O\sim P^{\prime }-O$ with $P\ne P^{\prime }$, a rational function with divisor $P-P^{\prime }$ would give a degree-one map to the projective line, contradicting that $E$ has genus one. The Abel--Jacobi map transports addition in the divisor class group to the elliptic-curve group law.

Given a base point $p_0$, the Abel map takes points of a smooth projective curve to its degree-zero Picard group of a curve. For positive genus it is injective: equality at distinct $p,q$ would give a principal divisor with one simple zero and one simple pole, forcing a degree-one map to the projective line and genus zero. For genus one the Riemann-Roch theorem gives a unique effective degree-one representative of $L(p_0)$ for every degree-zero line bundle $L$, making the map bijective and inducing an abelian group law on the points.

For a smooth genus one curve with origin $O$, the map $P\mapsto [(P)-(O)]$ identifies geometric points with the degree-zero Picard group of a curve. Riemann-Roch theorem gives a unique effective representative for each divisor class of degree one: its Riemann-Roch space has dimension one. A line cuts a plane cubic in a divisor linearly equivalent to $3(O)$ when $O$ is the Weierstrass flex. The two successive line intersections in the chord-and-tangent group law therefore give addition of the corresponding divisor classes. Associativity follows by injectivity of the identification and associativity of divisor-class addition, including tangent intersection multiplicities.

If $K_X=(\omega )$ is represented by a nonzero rational differential, multiplication by $\omega$ identifies the Riemann-Roch space $L(K_X)$ with the vector space of holomorphic differential forms. The Riemann-Roch theorem gives $\ell (K_X)=g$ for a smooth projective curve of genus $g$.

If a line bundle on a smooth projective curve has both $h^0(\mathcal{L})>0$ and $h^0({\omega }_C\otimes {\mathcal{L}}^{-1})>0$, the displayed inequality holds. Write these dimensions as $r,s$. The product dimension bound for sections on a curve gives $r+s-1\le g$, and the Riemann-Roch theorem gives $r-s=\mathrm{deg}\mathcal{L}+1-g$. Adding gives $2r\le \mathrm{deg}\mathcal{L}+2$.

In a proper flat family of algebraic curves, if the first sheaf cohomology of a line bundle vanishes on every fibre of an open set, its direct image there is a vector bundle of rank given by the Riemann-Roch theorem and commutes with arbitrary base change. Locally a finite free complex computes both fibre sheaf cohomology groups. Vanishing of the last cokernel makes its last map surjective; splitting that map leaves a finite locally free kernel in degree zero. This is the mechanism behind charts of the Picard scheme of a curve and evaluation descriptions of the theta divisor.

A Riemann surface which is compact as a topological space. For a connected one of genus $g$, the Riemann-Roch theorem gives $g$ independent holomorphic differential forms. The Riemann bilinear relations for a compact surface make their normalized periods a full lattice in ${\mathbb{C}}^g$, and the Abel-Jacobi map of a compact Riemann surface turns this lattice into a complex torus encoding divisor classes.

For a line bundle $M$ on $C\times T$, with $C$ a smooth projective curve, this function is locally constant. Twist both $M$ and $M^{-1}$ by a large power of an ample line bundle on $C$. Near a chosen fiber their first sheaf cohomology groups vanish. The Riemann-Roch theorem and upper semicontinuity of their zeroth sheaf cohomology groups then bound the degree from above and below by the degree on that fiber. Connectedness of $T$ is necessary for a single constant across all fibers.

For a torsion-free modular curve with regular modular cusps, let $f$ be a nonzero meromorphic weight-$k$ form and $C$ the reduced sum of modular cusp points. The local-order divisor of $f$, minus $C$, imposes exactly holomorphy in the interior and vanishing at every modular cusp on the product $f\phi$. Dividing any cusp form by $f$ gives the reverse identification with a Riemann-Roch space. The regular-cusp valence formula on a torsion-free modular curve and Riemann-Roch theorem compute the dimension when the resulting divisor has degree greater than the canonical degree.

Fix $O\in E(k)$. On a genus-one smooth projective curve, every degree-one divisor class has a unique effective point representative by the Riemann-Roch theorem: its section space has dimension one, since the complementary canonical-divisor degree is negative. Thus $P\mapsto [P-O]$ is a bijection onto the degree-zero Picard group of a curve. Transporting divisor-class addition proves the group axioms, with identity $O$, and characterizes $P+Q$ by $P+Q\sim R+O$.

Let $P=p_1+\cdots +p_k$ and $Q=q_1+\cdots +q_k$ be disjoint sums of distinct points on a smooth projective curve. A nonzero function with principal divisor $P-Q$ exists exactly when $L(Q-P)\ne 0$: a nonzero member makes $(f)+Q-P$ an effective divisor of degree zero, forcing equality. Evaluation $f\mapsto (f(p_1),\dots ,f(p_k))$ on $L(Q)$ has kernel of a linear map $L(Q-P)$.

If $\ell (Q)=k$, a choice of basis $f_1,\dots ,f_k$ turns the criterion into $\det (f_j(p_i))=0$. This determinant is not identically zero. Starting with $L(Q)$, repeatedly choose a point where some nonzero function in the current space does not vanish; evaluation then lowers the dimension of that space by one. After $k$ choices the joint kernel is zero. Thus a generic tuple of points gives an invertible evaluation matrix, and no function with divisor $P-Q$. On a genus one curve, Riemann-Roch theorem gives $\ell (Q)=k$, so this obstruction applies for every $k>0$.

A nonidentity holomorphic automorphism of a compact connected Riemann surface of genus $g$ has at most $2g+2$ fixed points. If there were more, choose $g+1$ distinct fixed points and let $D$ be their sum. The Riemann-Roch theorem gives $\ell (D)\ge 2$, so there is a nonconstant meromorphic function with at most simple poles at these points and no other poles. The function $h=f-f\circ \phi$ has at most $g+1$ poles and at least $n-g-1$ zeros at the remaining fixed points. Since a nonzero principal divisor has degree zero, $n>2g+2$ forces $h=0$. At an actual pole fixed by $\phi$, the reciprocal $1/f$ is a local coordinate because its zero is simple. The identity $f=f\circ \phi$ therefore forces $\phi$ to be the identity near that pole, and the identity theorem on a Riemann surface makes it the identity everywhere.

The embedding line bundle has at least four independent global sections. If it is nonspecial, the Riemann-Roch theorem gives $h^0=6-g$, so $g\le 2$. If it is special, the Clifford inequality for curves gives $h^0\le 3$, a contradiction. The smooth projective curve need not span its full ambient projective space.

For a smooth projective curve with $H^0(C,{\mathcal{O}}_C)=k$, its genus is ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$, equivalently ${\dim }_k{H}^0(C,{\Omega }_C^1)$ by Serre duality. It equals the arithmetic genus $1-\chi ({\mathcal{O}}_C)$. Without the condition on constants, the general Riemann-Roch theorem instead uses $\chi ({\mathcal{O}}_C)=h^0({\mathcal{O}}_C)-h^1({\mathcal{O}}_C)$.

On a smooth projective curve of genus two, $D=K+P+Q$ has three sections and no base points by the Riemann-Roch theorem. Its map to ${\mathbb{P}}^2$ is nondegenerate and has mapping degree times image degree four. The only nonbirational possibility is a double cover of a conic. Every degree-two divisor with two sections is canonical, again by Riemann-Roch theorem, so that possibility forces $D\sim 2K$, or $P+Q\sim K$. If this equivalence fails, the image is a quartic and the map is birational. The quartic is singular because a smooth plane quartic has genus three.

A weight-$2k$ modular form descends as an invariant tensor differential. At an elliptic point of order $e$, a local invariant coordinate is $w=u^e$, so its descended coefficient has a pole of order at most $\lfloor k(e-1)/e\rfloor$. At infinity, $d\tau =dq/(2\pi iq)$ gives a pole of order at most $k$. Thus $D=k[\infty ]+\lfloor k/2\rfloor [i]+\lfloor 2k/3\rfloor [\rho ]$ bounds the poles. Conversely these bounds give holomorphic pullbacks. The Riemann-Roch theorem on the compactified modular curve proves finite-dimensionality; since $X(1)$ is the Riemann sphere, the dimension is $\max (0,1-k+\lfloor k/2\rfloor +\lfloor 2k/3\rfloor )$.

If a divisor on an algebraic curve $D$ on a smooth projective curve has negative degree of a divisor, then its Riemann-Roch space is zero. Otherwise a nonzero $f\in L(D)$ would make $(f)+D$ an effective divisor of degree $\mathrm{deg}D<0$, since a principal divisor has degree zero. This is impossible. In genus one the Riemann-Roch theorem consequently gives $\ell (D)=\mathrm{deg}D$ for every divisor of positive degree.

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.

Choose a nonzero meromorphic differential on a Riemann surface $\eta$ and write $K=(\eta )$. Multiplication defines a linear mapIt takes values in the holomorphic differential forms because $(f\eta )=(f)+K\ge 0$. It is injective since $\eta \ne 0$. It is surjective because, for any holomorphic differential form $\omega$, the ratio $\omega /\eta$ is a meromorphic function and has $(\omega /\eta )+K=(\omega )\ge 0$. Thus this map identifies $\mathrm{HD}(C)$ with the canonical Riemann-Roch space.

Furthermore, $\mathcal{L}(0)=\mathbb{C}$: a meromorphic function with no poles is holomorphic on the compact Riemann surface $C$, and hence constant by the maximum modulus principle. Applying the Riemann-Roch theorem with $D=0$ givesTherefore

First fix distinct $q_1,\dots ,q_k$ and let $Q=q_1+\cdots +q_k$. A negative-degree divisor has no sections: if $0\ne f\in \mathcal{L}(D)$, then $(f)+D$ is an effective divisor of degree $\mathrm{deg}D$, since a principal divisor has degree zero, so $\mathrm{deg}D$ cannot be negative. As $\mathrm{deg}K=0$ and $\mathrm{deg}(K-Q)=-k<0$, the Riemann-Roch theorem therefore gives

Now choose distinct $p_1,\dots ,p_k$ outside the support of $Q$, and put $P=p_1+\cdots +p_k$. A function with the required zeros and poles must be a nonzero element of $\mathcal{L}(Q-P)$. Conversely, if $0\ne f\in \mathcal{L}(Q-P)$, then $(f)+Q-P$ is an effective divisor of degree zero, hence is zero. It follows that $(f)=P-Q$, giving precisely the required simple zeros and simple poles, with no others. Thus the existence question is exactly whether $\mathcal{L}(Q-P)\ne 0$.

Take a basis $f_1,\dots ,f_k$ of the Riemann-Roch space $\mathcal{L}(Q)$. Evaluation at the $p_i$ defines the linear mapwhose kernel is $\mathcal{L}(Q-P)$. The evaluation criterion for a prescribed principal divisor becomesThe crucial point is that this determinant does not vanish identically; a count of unknowns and equations alone would not prove this.

To verify it, begin with $V_0=\mathcal{L}(Q)$, of dimension $k$. Whenever $V_{j-1}\ne 0$, choose a nonzero $f\in V_{j-1}$. Its zeros are finite. Choose $p_j$ outside those zeros, the support of $Q$ and the previously chosen points. Evaluation at $p_j$ is then a nonzero linear functional on $V_{j-1}$, soAfter $k$ steps, $V_k=0$. Thus there exists a tuple of distinct points with $\Delta \ne 0$.

The entries of this determinant vary holomorphically away from the $q_i$, and are rational functions on the nonsingular projective curve. Its vanishing is therefore a proper closed exceptional locus on this configuration space; its complement is open and dense in the Zariski topology. For a generic choice of the $p_i$, $E_P$ is invertible andThis holds for every fixed tuple of distinct $q_i$, which proves the assertion for generic choices of both sets of points. At $k=1$, it is especially transparent: $\mathcal{L}(q_1)=\mathbb{C}$, so no nonzero member can vanish at a distinct $p_1$.