Large Intersection Multiplicity

Question

A cubic curve, say, $x^3+y^3=1$ and some quadratic curve $f(x,y)=0$ generally have six intersection points in $\mathbb{CP}^2$.

Question: If all the intersection points coincide, what will be the exact form of $f(x,y)$?

Answer 1

This is a partial answer, which expands my comment above. It uses crucially that your cubic curve can be given the structure of an elliptic curve. I will quickly recall what that means. The reference is Fulton, Chapter 5.

Let $C \subset \Bbb{P}^2$ be a non-singular, projective plane cubic curve. After choosing a point $O \in C$, we can define an addition law on $C$, which turns $C$ into an abelian group. The pair $(C,O)$ is called an elliptic curve (with the addition law induced by $O$ understood). So an elliptic curve is a non-singular cubic curve, which is also an abelian group. In particular, it makes sense to talk about the order of a point $P \in C$.

A flex on $C$ is a point $P \in C$, such that the tangent line to $C$ at $P$ intersects $C$ only at $P$. One can show that every non-singular cubic has a flex (at least in characteristic zero).

Why all this talk about elliptic curves and flexes? Because of the following theorem. Let $(C,O)$ be an elliptic curve, with $O \in C$ a flex, and $P \in C$.

Theorem: There exists an irreducible quadratic curve $C' \subset \Bbb{P}^2$, which intersects $C$ only at $P$, if and only if $6P=O$ and $3P \neq O$.

For example, $P \neq O$ could be a point of order two in $(C,O)$. The Theorem itself follows from the more general Problem 5.41 of Fulton, and is a nice application of Max Noether's AF+BG Theorem (cf. Fulton, Chapter 5.5).

Now let $C \subset \Bbb{P}^2$ be the curve defined by $X^3+Y^3-Z^3=0$ (i.e. the homogenization of your curve above). The point $O:=[1:0:1]$ is a flex on $C$. Also one can show that the point $P:=[-1:\sqrt[3]{2}:1]$ has order two in $(C,O)$. Hence by the above theorem, there exists an irreducible quadratic curve $C'$, which intersects $C$ only at $P$. One such curve is given (if I made no mistakes) by the quadratic form

$$ F(X,Y,Z)=-3X^2+3\sqrt[3]{4}Y^2+3\sqrt[3]{2}XY-6XZ-3 \sqrt[3]{2}YZ-3Z^2 \in \Bbb{C}[X,Y,Z] $$

Finding such $F$ explicitly is, again, an application of the AF+BG Theorem.

I can give some more details, if you wish so, but I feel that I should leave it at that for the moment.

Answer 2

I haven't ever seen a proof of this, but it seems that curves that intersect with multiplicity 2 or higher have the same tangent line, and multiplicity 3 or higher have same second derivative and so on.

From that, the only points where a quadratic curve could have a intersection of higher multiplicity than 4 with a cubic curve are the points of the cubic curve with a vanishing third derivative, because a quadratic curve always has a vanishing third derivative.

Finding the third derivative of x^3+y^3-1=0 is difficult, but with some approximation algorithms I find that the quadratic curve 0.1893004115226*x^2+0.3127572016461*x+2.8683127572018=y approximately intersects the curve x^3+y^3-18=0 with a multiplicity of at least 4 at a point near (-2,3).