= Solution
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> write
$$
D=\sum_p n_pp,\qquad \sum_p n_p=0,
$$
since every closed point is a $k$-point. Therefore
$$
D=\sum_p n_p(p-p_0),\qquad
\mathcal O(D)\cong\bigotimes_p\mathcal O(p-p_0)^{\otimes n_p}.
$$
Only finitely many factors occur, and negative tensor powers mean dual <line bundles>. This proves
$$
\boxed{\operatorname{Pic}^0(X)=\langle\alpha(X(k))\rangle.}
$$
For 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 \b[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 equation
$$
y^2+y=x^5+x+1.
$$
The 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 $2\deg A$ and $2\deg B+5$, of opposite parity, so they cannot cancel. The pole bound forces $B=0$ and $\deg A\le1$. 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.
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