Divisors on hyperelliptic curves

Abstract

Given a finite number of distinct elements a_i ∈ ¢, i ∈ S, let $f\left( t \right) = \prod\limits_{i \in S} {\left( {t - a_i } \right)}$. We form the plane curve C₁ defined by the equation

$$ s^2 = f\left( t \right). $$

The polynomial s² -f(t) is irreducible, so (s² -f(t)) is a prime ideal, and C₁ is a 1-dimensional affine variety in ¢². In fact, C₁ is smooth. To prove this, we will calculate the dimension of the Zariski-tangent space at each point, i.e., the space of solutions $(\dot s,\dot t)$ ∈ ¢² to the equation

$$ \left( {s + \varepsilon \dot s} \right)^2 \equiv \prod {\left( {t + \varepsilon \dot t - a_i } \right)} \bmod \varepsilon ^2 for\left( {s,t} \right) \in c_1 . $$

That is equivalent to the equation

$$ 2s\dot s = \dot t \cdot \sum\limits_{j \in S} {\prod\limits_{i \ne j} {\left( {t - a_i } \right)} } ; $$

if s ≠ 0, the solutions are all linearly dependent since $\dot s = \frac{{\dot t}} {{2s}}\left( {\sum\limits_{j \in S} {\prod\limits_{i \ne j} {\left( {t - a_i } \right)} } } \right)$ ; if s = 0, we get from the equation of the curve $\prod\limits_{i \in S} {\left( {t - a_i } \right)} = 0$, hence t = a_i for some i; thus $0 = \dot t \cdot \prod\limits_{j \ne i} {\left( {a_i - a_j } \right)}$, so $\dot t = 0$. Thus at all points, the Zariski tangent space is one-dimensional.