Principal divisor on an algebraic curve ($(f)$)

For a nonzero rational function $f$ on a smooth projective curve $C$, its principal divisor isIt records the zeros of $f$ with positive multiplicity and its poles with negative multiplicity. Every principal divisor has degree zero.

For $E=\mathbb{C}/\Lambda$, integrate the logarithmic derivative of an elliptic function and its product with $z$ around a fundamental cell. The first integral forces total degree zero; the second forces the weighted point sum to lie in the lattice. Conversely choose lifts of the points with equal zero and pole sums, and form the balanced ratio of translated Weierstrass sigma functions. Their quasi-period factors cancel, giving an elliptic function with exactly the prescribed divisor. Separate zero and pole divisors require disjoint support; common terms cancel when only their difference is prescribed.

Every degree-zero divisor on the projective line is principal. Indeed, if $D={\sum }_i{a}_i{p}_i$ and $L_i$ is a homogeneous linear form vanishing at $p_i$, then ${\sum }_i{a}_i=0$ makesa degree-zero rational function with $(f)=D$. Consequently, two divisors on ${\mathbb{P}}^1$ are linearly equivalent exactly when they have the same degree.

If distinct points $p,q$ on a smooth projective curve satisfy $(f)=p-q$, then $f:C\to {\mathbb{P}}^1$ is a finite morphism of degree one. Such a morphism between smooth projective curves is an isomorphism, so $C$ has genus zero. Hence no divisor $p-q$ on a curve of positive genus is principal.

For an elliptic curve with identity $O$, a divisor $D={\sum }_P{n}_P(P)$ is principal exactly whenThis is the identification of the degree-zero divisor class group with the elliptic curve.