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 ...

Get Discrete Algebraic Methods now with the O’Reilly learning platform.

O’Reilly members experience books, live events, courses curated by job role, and more from O’Reilly and nearly 200 top publishers.