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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.