For which points $P$ on a nonsingular cubic $C$ does there exist a nonsingular conic that intersects $C$ only at $P$?

Question

For which points $P$ on a nonsingular cubic $C$ does there exist a nonsingular conic that intersects $C$ only at $P$?

By Bezout's theorem, we must have that $I(P,C \cap F)=6$, where $F$ is the required conic. But I can't really do anything with this information. Can anyone give any hints?

Answer 1

Short answer: homogeneous degree-2 forms give (all) the sections of $\mathcal{O}_C(6p)$ where $p$ is any inflection point. This follows from the fact that $\mathcal{O}(1)|_C \cong \mathcal{O}_C(3p)$ and the fact that $C$ is projectively normal.

So, such a form $F$ can have $\mathrm{div}(F) = 6q$ if and only if $6p = 6q$ in $\mathrm{Pic}(C)$, that is, $q$ is $p$ plus a $6$-torsion point. There are 36 such points.

Longer answer: I got kind of nerd-sniped by this and figured out a (sorta ridiculous) direct construction. Here it is. Assume the curve is in Weierstrass form, so "vertical line through $p$" means "line through $p$ and the point at infinity $p_\infty$", and also $p_\infty$ is an inflection point. Let:

• $p_1 \in C$ be a torsion point of order 6.
• $T$ be its tangent line and let $p_4$ be the third point of $T \cap C$.
• $V$ be the vertical line through $p_4$ and let $p_2$ be the third point of $V \cap C$.
• $L$ be the line through $p_1$ and $p_2$ and let $p_3$ be the third point of $L \cap C$.
• $V'$ be the vertical line through $p_3$.
• $T_\infty$ be the tangent line through the point at infinity.

Then one checks that $$\mathrm{div} \bigg(\frac{T^2 L^2 T_\infty}{V^2 V'} \bigg) = 6[p_1],$$ so in particular this form is regular on $C$. Therefore it comes from a degree-$2$ polynomial $F$ on $\mathbb{P}^2$ (this uses projective normality to say that every section of $\mathcal{O}(6p)$ comes from $\mathbb{P}^2$).

Proof: I messed around with the addition law for elliptic curves until I found the formula above.