Jacobians of Curves

For those familiar with algebraic geometry and algebraic curves, one can prove Theorem 19.2.3 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.

$\displaystyle 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.

William Stein 2012-09-24