Discrete Algebraic Methods

Let us first consider the class of principal divisors. There is no other divisor in this class: suppose that D + div(f) = div(h) for polynomials f, h ∈ k[x, y]. Then, ordP(f) ≤ ordP(h) for all P ∈ E (k) and, by Theorem 5.4, there exists a polynomial g with fg = h. Thus, D = div(g) is a principal divisor. In particular, the zero class does not contain any divisor of degree 1. For P ∈ E(k), it then follows that [P] ≠ 0 in Pic0(E(k)) and therefore the Picard group is not trivial.

If we consider two points P, Q ∈ E(k), then [P] ≠ [Q] is equivalent to If the x-coordinates of P and are different and there is a unique line through P and This line ...