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Nov 17, 2011 at 3:57 comment added solbap I think its worth noting that you can still define the group law even when you choose a point $O$ that is not an inflection point. Thinking of $E \subset \mathbb{P}^2$, if $P,Q \in E$ let $l_{P,Q}$ be the line joining them. Let $R$ be the third point of intersection in $E \cap l_{P,Q}$ then $P+Q \in E$ is the third point of intersection in $E \cap l_{O,R}$. This works because the divisor of $l_{P,Q}/l_{O,R}$ is $P + Q - (P+Q) - O$. In other words, the line bundle $O_E(P - O)\otimes O_E(Q - O)$ is isomorphic to $O_E((P+Q)- O)$.
Nov 17, 2011 at 3:27 comment added Joe Silverman @Sasha: good point, it only works if $O$ is an inflexion point.
Nov 17, 2011 at 3:26 vote accept unknown
Nov 17, 2011 at 3:23 history answered Sasha CC BY-SA 3.0