Divisors on hyperelliptic curves

Abstract

Given a finite number of distinct elements ai ∈ ¢, 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 = ai 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.