Third point of intersection is also a point of inflection?

Question

Let $C \subset \mathbb{P}_2$ be a nonsingular cubic. If $L$ is a line through two distinct points of inflection on $C$, how do I show that the third point of intersection is also a point of inflection?

Answer 1

Hint. Let $p, q \in C$ be two points of inflection, and let $L$ be the line between them. Let $L \cdot C = p + q + r$. Let $M$ the tangent line to $C$ at $r$, and $M \cdot C = 2r + s$. Show that $r = s$ by showing that $s$ lies on the line through $p$ and $q$.

---

In the interests of completeness, we provide a complete solution. Let $M_1$ be the line tangent to $C$ at $p$, and $M_2$ the line tangent to $C$ at $q$. Then $M_1 \cdot C = 3p$ and $M_2 \cdot C = 3q$. Let $D = M \cup M_1 \cup M_2$. Then$$D \cdot C = 3p + 3q + 2r + s \ge 2(L \cdot C).$$By Noether's Theorem, there exists a curve $E$ of degree at most $\text{deg}(D) - \text{deg}(2L) = 1$ such that $E \cdot C = p + q + s$. We have that $E$ is the line through $p$ and $q$, so $E = L$ and $r = s$, so $M \cdot C = 3r$. Thus, $r$ is a point of inflection.

For a different proof, see Theorem 6 here.

Answer 2

A less direct explanation than that of Kevin Dong: your curve is elliptic, and any point of inflection may be taken to be the identity $\Bbb O$, after which choice you can use chord-and-tangent to describe the addition. Then, any point of inflection is a $3$-torsion point, and the set of all these is a group.

Answer 3

I've found in Salmon -- A treatise on the higher plane curves, a proof of the following fact:

If $a$, $b$, $c$ are real inflection points of a real cubic curve, then $a$, $b$, $c$ are collinear.

Proof: draw the tangents to the curve at points $a$, $b$, $c$, forming a triangle $ABC$ ( $a$ on $BC$, etc). Now apply Carnot's theorem to the points $A$, $B$, $C$. We have

$$\frac{Ba^3}{Ca^3} \cdot \frac{Cb^3}{Ab^3} \cdot \frac{Ac^3}{Bc^3} = 1$$

(the line $AB$ has triple contact with the curve at $c$, etc). Now, since we are dealing with real quantities, we conclude that $$\frac{Ba}{Ca} \cdot \frac{Cb}{Ab} \cdot \frac{Ac}{Bc} = 1$$ so by Menelaus, the points $a$, $b$, $c$ are collinear.

With this idea, let's prove that the line through two inflection points $a$, $b$ intersects the cubic in a third inflection points (($\operatorname{char} k, 6)=1$). For let $c$ be the third intersection of the line with the cubic. Again, draw the tangents to the cubic at $a$, $b$, $c$ forming like before the triangle $ABC$. Note that apriori the tangent at $c$ intersects the curve in a third point $c'$, and we have to show that $c'=c$. Apply again Carnot:

$$\frac{Ba^3}{Ca^3} \cdot \frac{Cb^3}{Ab^3} \cdot \frac{Ac^2}{Bc^2}\cdot \frac{A c'}{Bc'} = 1$$ and also take Menelaus$\ ^3$. We conclude $$\frac{Ac}{Bc} = \frac{Ac'}{Bc'}$$ and so $c=c'$.

$\bf{Added:}$ In a similar way we can prove a particular case of Cayley-Bacharach theorem:

Let $ABC$ a triangle intersecting a cubic in $9$ points $a_1$, $a_2$ $a_3$ ( on $BC$), $\ldots$, $c_1$, $c_2$, $c_3$. If another cubic passes through $8$ of the $9$ points $a_i$, $b_i$, $c_i$, then it also passes through the ninth.

Indeed, let $c_3'$ the ninth point of intersection of $ABC$ with the second cubic. Write the Carnot equality for the triangle $ABC$ the each cubic $$\prod\frac{Ac_1\cdot Ac_2 \cdot Ac_3}{Bc_1 \cdot Bc_2 \cdot Bc_3} = 1$$ $$\ldots$$ and get from the above $$\frac{Ac_3}{Bc_3} = \frac{Ac_3'}{Bc_3'}$$ so $c_3' = c_3$.