Degree-zero Picard group of a curve (${\mathrm{Pic}}^0(C)$)

For a smooth projective curve, the degree-zero Picard group consists of line bundles of degree zero up to isomorphism, with tensor product as addition. A nonzero rational section of a line bundle identifies each line bundle with $\mathcal{O}(D)$ for a divisor on an algebraic curve; degree is the sum of multiplicities weighted by residue-field degrees. Over an algebraically closed field, every degree-zero divisor is a sum of differences of points.

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.

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.