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Hi,

Using the isomorphism between an elliptic curve $E$ and its $Pic_1(E)$ group, one can easily give $E$ the structure of a group variety after choosing a point $O\in E$. The operation that one gets is: $$P+Q+R = 0\text{ in $E$ iff }P+Q+R-3O\text{ (as divisors) is a principal divisor}.$$

  • Question: Why is the condition $P+Q+R-3O$ being a principal divisor equivalent to $P+Q+R$ being the intersection divisor of $E$ with a line?

Thanks!

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This is true if $O$ is an inflexion point. In this case $3O$ is cut out by a line, so $3O$ is linearly equivalent to the intersection of a line with a curve. Thus $P + Q + R$ is linearly equivalent to $3O$ if and only if it is cut out by a line.

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  • $\begingroup$ @Sasha: good point, it only works if $O$ is an inflexion point. $\endgroup$
    – Joe Silverman
    Commented Nov 17, 2011 at 3:27
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    $\begingroup$ 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 $O_E((P+Q)- O)$. $\endgroup$
    – solbap
    Commented Nov 17, 2011 at 3:57
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Rough idea (you can fill in the details): If $P+Q+R-3O$ is a principal divisor, it's the divisor of $f(x,y)$, and $f=0$ intersects $E$ in three points iff $f$ is a linear function, so $P,Q,R$ lie on a line. Converse is pretty clear.

This is a pretty elementary question, did you really spend a significant amount of time thinking about it before posting it on MO? If not, you're not using MO properly.

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    $\begingroup$ But if "unknown" did spend time thinking about it, then he or she may have been made to feel stupid for asking. One person's difficulty is another person's triviality. $\endgroup$
    – Todd Trimble
    Commented Nov 17, 2011 at 14:32

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