If a plane curve of degree n intersects an elliptic curve in 3n points, then do those points always sum to zero when added using the group law on the points of an elliptic curve ?
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asked Nov 1, 2011 at 17:08
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2$\begingroup$ Denote by p your favorite flex on the plane cubic E, by H the line through p, and by C the degree n curve, then CE ~ nHE because C~nH in the plane. $\endgroup$– David LehaviCommented Nov 1, 2011 at 17:40
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2$\begingroup$ Yes, they do... $\endgroup$– Felipe VolochCommented Nov 1, 2011 at 17:41
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5$\begingroup$ I'm not sure what qualifies as a "homework level problem", but it seems to me that both this one and the previous one that you posted (about intersection of a quartic and a conic) fall into that category. If I assigned them in a grad class, I'd certainly expect everyone to be able to solve them. Did you actually spend several hours working on these problems and make no progress? I'm happy with Felipe's answer of a bald "Yes", but really, people, if a problem is this simple, suggest the poster work on it some more instead of answering. Or at most, give a hint. $\endgroup$– Joe SilvermanCommented Nov 1, 2011 at 20:39
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2 Answers
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I assume that elliptic curve means a cubic curve $E$ with some base point $O$.
Then the answer to your question is yes if and only if the base point O for the group law is a flex point of $E$. (In many texts it is assumed that the base point of $E$ is a flex line, but to define the group law this is completely unnecessary, see Fulton's book on plane curves for the geometric definition of the group law in this case.) Fulton's geometric definition of the group law on a cubic is equivalent with
$P_1,..P_{3n}$ add up to zero if and only if $P_1+..P_{3n}-3n O$ is zero in $Pic^0(E)$, i.e., is a divisor of a function on $E$.
Assume first that $O$ is a flex point. Let g be polynomial defining the flex line, let $f$ be a polynomial defining $C$ and let $P_1,...P_{3n}$ be the $3n$ intersection points of $C$ and $E$. As David commented above we have $div(f/g^n)=P_1+..P_{3n}-3nO$, hence the points add up to zero.
Suppose now that O is not a flex. Let $R$ be the third intersection point of $E$ with the tangent line of $E$ at $O$. Since $O$ is not a flex we have that $R\neq O$. To check whether these points add up to zero we have to calculate $O-O+O-O+R-O$ in $Pic^0(C)$. It is easy to see that this divisor is nonzero in $Pic^0(C)$.
answered Nov 1, 2011 at 19:17
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2$\begingroup$ Can we say the converse? i.e., if $P_1\oplus...\oplus P_{3n}=O$ in E (with flex O), then there is a curve $C$ with degree $n$ through $P_1,...,P_{3n}$? $\endgroup$– rmdmc89Commented Aug 6, 2021 at 19:11
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$\begingroup$ @rmdmc89: Vous devriez poser votre question en tant que question indépendante sur ce site: votre commentaire ci-dessus n'a aucune visibilité. $\endgroup$ Commented Feb 22, 2022 at 22:22
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$\begingroup$ En attendant , la réponse est oui: par Abel-Jacobi nous avons $ P_1+...+P_{3n}\equiv 3d\cdot \mathcal O\in Div(E)$, la congruence étant prise au sens de l'équivalence linéaire des diviseurs de $E$. Comme la courbe $E$ est projectivement normale (toutes les courbes planes le sont!) le système $\vert \mathcal O(n)\vert$ est complet et on a bien $P_1+...+P_{3n}=C\cdot E$ pour une courbe $C\subset \mathbb P^2$ convenable. $\endgroup$ Commented Feb 23, 2022 at 9:11
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Yes.
Choose a flex as zero point. The sum of three points on a line is zero essentially by construction of the group law. Now the divisor cut on the elliptic curve by a curve of degree $n$ is linearly equivalent to the sum of $n$ hyperplane divisors, so the claim follows.
answered Nov 1, 2011 at 17:41