Third point of intersection is also a point of inflection? 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?
 A: 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.
A: 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.
A: This is Problem 2 in section 5.7, "Elliptic Curves," of I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991.  I use the nomenclature from that book.
After I worked out this answer, I realized that it is the same as Lubin's.
$C$ is elliptic.  Let $\mathbf{A}$ and $\mathbf{B}$ be the two distinct inflection points and $\mathbf{AB}$ the third point.  Choose $\mathbf{0}$ to be an inflection point.  We use the following facts.  The proof of the main result is farther down.


*

*Fact 1: $\mathbf{AA} = \mathbf{A}$, $\mathbf{BB} = \mathbf{B}$, and $\mathbf{00} = \mathbf{0}$.

*Converse of Fact 1:  If $\mathbf{PP} = \mathbf{P}$, then $\mathbf{P}$ is an inflection point.

*Fact 2:  $\mathbf{P}(\mathbf{0P}) = \mathbf{0}$.  Indeed, $\mathbf{0P}$ is the third point on the line through $\mathbf{0}$ and $\mathbf{P}$.  The line through $\mathbf{P}$ and $\mathbf{0P}$ is the same line, so the third point on the line is the original point $\mathbf{0}$.

*Fact 3: $\mathbf{A} + \mathbf{A} + \mathbf{A} = \mathbf{0}$, proved as follows:
\begin{align}
\mathbf{A} + \mathbf{A} + \mathbf{A} & = \mathbf{A} + \mathbf{0(AA)} &&\text{definition of +}\\
& = \mathbf{A} + \mathbf{0(A)} &&\text{Fact 1}\\
& = \mathbf{0(A(0A))} &&\text{definition of +}\\
& = \mathbf{00} &&\text{Fact 2}\\
& = \mathbf{0} &&\text{Fact 1}
\end{align}
Likewise, $\mathbf{B} + \mathbf{B} + \mathbf{B} = \mathbf{0}$.

*Converse of Fact 3:  If $\mathbf{P} + \mathbf{P} + \mathbf{P} = \mathbf{0}$, then $\mathbf{P}$ is an inflection point.  To prove that, we  have
\begin{align}
\mathbf{P} + \mathbf{P} + \mathbf{P} & = \mathbf{0}\\
\mathbf{0(P(0(PP)))} & = \mathbf{00} &&\text{definition of + on left; Fact 1 on right}\\
\mathbf{P(0(PP))} & = \mathbf{0} &&\text{by uniqueness}
\end{align}
We also know that $\mathbf{P(0P)} =  \mathbf{0}$ by Fact 2.  Comparing these last two equations, we see that $\mathbf{PP} = \mathbf{P}$ by uniqueness.  Hence, $\mathbf{P}$ is an inflection point by the converse of Fact 1.

*Fact 4: $\mathbf{A} + \mathbf{B} + \mathbf{AB} = \mathbf{0}$ is proved at The sum of three colinear rational points is equal to $O$.  (That proof applies to general points, not necessarily rational.)
We are now ready for the proof of the main result:
\begin{align}
\mathbf{0} & = \mathbf{A} + \mathbf{A} + \mathbf{A} + \mathbf{B} + \mathbf{B} + \mathbf{B} &&\text{Fact 3}\\
& = \mathbf{A} + \mathbf{B} + \mathbf{A} + \mathbf{B} + \mathbf{A} + \mathbf{B} &&\text{+ is commutative}\\
& = -\mathbf{AB} - \mathbf{AB} - \mathbf{AB} &&\text{Fact 4}
\end{align}
Because $-\mathbf{0} = \mathbf{0}$, we have that $\mathbf{AB} + \mathbf{AB} + \mathbf{AB} = \mathbf{0}$.  Thus, by the converse of Fact 3, $\mathbf{AB}$ is an inflection point.
A: 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 $c_3' = c_3$.
