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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.

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 $E$ 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.

Here the degree-zero Picard group of a curve uses tensor product as its group operation. The map is the Abel map of a pointed smooth projective curve. If $\alpha (p)=\alpha (q)$, then $\mathcal{O}(p-q)$ is trivial. By the divisor description of line bundles, there is a nonzero rational function $f$ with principal divisor $(f)=p-q$.

If $p\ne q$, this function has exactly one pole, of order one. It defines a nonconstant morphism of algebraic varieties $X\to {\mathbb{P}}^1$: locally at a pole one uses $1/f$ as the target coordinate. The degree of a morphism of curves is the total pole degree, so it is one. The map is finite, since a nonconstant map of projective integral curves has finite fibers and is proper, and a degree-one finite map to the normal projective line is an isomorphism. Algebraically, on a target affine chart the finite coordinate algebra lies inside the common function field and is integral over the integrally closed coordinate ring, so it equals that ring. Thus $X\cong {\mathbb{P}}^1$ and its genus is zero, contradicting $g\ge 1$. ConsequentlyThis spells out why a principal divisor with one simple zero and one simple pole cannot occur on a curve of positive genus.

If $g=0$, the chosen point $p_0$ is enough to identify $X$ with ${\mathbb{P}}^1$, even over a nonclosed field. By the Riemann-Roch theorem, $h^0(\mathcal{O}(p_0))=2$, since the complementary canonical twist has degree $-3$ and no nonzero section. Besides constants there is a rational function with its sole pole a simple pole at $p_0$, giving the same degree-one isomorphism. On ${\mathbb{P}}^1$, every degree-zero divisor is principal: over an algebraically closed field write a product of powers of linear forms for its finite point coefficients, with the degree-zero condition giving the coefficient at infinity. Over a general field the same construction uses irreducible polynomial factors for closed points, weighted by their degrees. Therefore

First use the algebraically closed field convention. Every line bundle on a smooth projective curve has a nonzero rational section of a line bundle: trivialize it on a nonempty open set and take its generic generator. The orders of its zeros and poles give a finite divisor $D$, and the local trivializations identify the line bundle with $\mathcal{O}(D)$. For a degree-zero line bundle writesince every closed point is a $k$-point. ThereforeOnly finitely many factors occur, and negative tensor powers mean dual line bundles. This provesFor genus zero the group is trivial, so the conclusion remains valid.

For a general field, a closed point need not be rational, and its contribution to degree is weighted by its residue degree. Consequently rational points need not generate the degree-zero Picard group of a curve. Here is an explicit counterexample to that stronger reading. Let $C$ be the smooth projective curve over ${\mathbb{F}}_2$ obtained by completing the affine equationThe polynomial is irreducible over ${\overline{\mathbb{F}}}_2(x)$: a function $r^2+r$ has even pole order at infinity when it has a pole there, while the right side has pole order five. Its affine model is smooth because the partial derivative with respect to $y$ is one. The normalized projective completion is therefore a geometrically integral smooth projective curve.

There is exactly one point $p_{\infty }$ above infinity, and it is rational. Indeed the map $x:C\to {\mathbb{P}}^1$ has degree two. At a point above infinity put $v(x)=-e$, where the ramification index $e$ is at most two. The equation forces $v(y)<0$ and $2v(y)=5v(x)=-5e$. Hence $e=2$ and $v(y)=-5$. The degree formula then leaves exactly one point of residue degree one. Thus $x$ and $y$ have pole orders two and five there. There are no affine rational points: for either $x\in {\mathbb{F}}_2$, the right side is one, while $y^2+y=0$ for every $y\in {\mathbb{F}}_2$. Hence $C({\mathbb{F}}_2)=\{p_{\infty }\}$ and the image of its Abel map of a pointed smooth projective curve is trivial.

The equations $y=0$, $x^2+x+1=0$ define a degree-two closed point $D$, since $x^5+x+1$ vanishes modulo $x^2+x+1$. The divisor $D-2p_{\infty }$ has degree zero but is not principal. To prove this, a function with poles bounded by $2p_{\infty }$ is regular on the affine curve and hence has a unique expression $A(x)+B(x)y$. The possible pole orders of its two terms are $2\mathrm{deg}A$ and $2\mathrm{deg}B+5$, of opposite parity, so they cannot cancel. The pole bound forces $B=0$ and $\mathrm{deg}A\le 1$. But no nonzero $a+bx$ over ${\mathbb{F}}_2$ vanishes at $D$, whose residue class of $x$ has irreducible quadratic minimal polynomial. Thus no function has principal divisor $D-2p_{\infty }$, proving its class is nonzero. This establishes the rational points need not generate the degree-zero Picard group qualification with a complete obstruction, while the generation proof above supplies the intended algebraically closed-field answer.

Let $C$ be the smooth projective model over ${\mathbb{F}}_2$ of $y^2+y=x^5+x+1$. Its only rational point is its unique point $p_{\infty }$ at infinity. At that point $x,y$ have pole orders two and five. The degree-two closed point $D$ defined by $y=0$, $x^2+x+1=0$ gives a nonzero class $[D-2p_{\infty }]$. Indeed a function with poles bounded by $2p_{\infty }$ has the form $a+bx$: in the affine coordinate ring every function is uniquely $A(x)+B(x)y$, and the two possible pole orders have opposite parity. No nonzero $a+bx$ vanishes at both roots of $x^2+x+1$. Thus the Abel map of a pointed smooth projective curve on rational points has trivial image here, but the degree-zero Picard group of a curve is nontrivial. Generation by point differences needs an algebraically closed field, or closed points with their residue degrees instead of just rational points.