Jacobians of Curves

For those familiar with algebraic geometry and algebraic curves, one can prove Theorem 21.2.3 from an alternative point of view. There is a bijection between nonsingular geometrically irreducible projective curves over $\mathbf{F}$ and function fields $K$ over $\mathbf{F}$ (which we assume are finite separable extensions of $\mathbf{F}(t)$ such that $\overline{\mathbf{F}}\cap K = \mathbf{F}$). Let $X$ be the curve corresponding to $K$. The group $D_K^0$ is in bijection with the divisors of degree 0 on $X$, a group typically denoted $\Div ^0(X)$. The quotient of $\Div ^0(X)$ by principal divisors is denoted $\Pic ^0(X)$. The Jacobian of $X$ is an abelian variety $J=\Jac (X)$ over the finite field $\mathbf{F}$ whose dimension is equal to the genus of $X$. Moreover, assuming $X$ has an $\mathbf{F}$-rational point, the elements of $\Pic ^0(X)$ are in natural bijection with the $\mathbf{F}$-rational points on $J$. In particular, with these hypothesis, the class group of $K$, which is isomorphic to $\Pic ^0(X)$, is in bijection with the group of $\mathbf{F}$-rational points on an algebraic variety over a finite field. This gives an alternative more complicated proof of finiteness of the degree 0 class group of a function field.

Without the degree 0 condition, the divisor class group won't be finite. It is an extension of  $\mathbf{Z}$ by a finite group.

$$0 \to \Pic ^0(X) \to \Pic (X) \xrightarrow{\deg} n\mathbf{Z}\to 0,$$

where $n$ is the greatest common divisor of the degrees of elements of $\Pic (X)$, which is $1$ when $X$ has a rational point.