The following exercise from Rational Points on Elliptic Curves asks to prove the cubic Cayley-Bacharach theorem in the case that eight points are distinct.
A.17. In this exercise we guide you in proving the cubic Cayley-Bacharach theorem in the case that the eight points are distinct. Let $C_1: F_1=0$ and $C_2$ : $F_2=0$ be cubic curves in $\mathbb{P}^2$ without common component which have eight distinct points $P_1, P_2, \ldots, P_8$ in common. Suppose that $C_3: F_3=0$ is a third cubic curve passing through these same eight points. Prove that $C_3$ is on the "line of cubics" joining $C_1$ and $C_2$, i.e., prove that there are constants $\lambda_1$ and $\lambda_2$ such that $$ F_3=\lambda_1 F_1+\lambda_2 F_2 $$ In order to prove this result, assume that no such $\lambda_1, \lambda_2$ exist and derive a contradiction as follows:
- (i) Show that $F_1, F_2$, and $F_3$ are linearly independent.
- (ii) Let $P^{\prime}$ and $P^{\prime \prime}$ be any two points in $\mathbb{P}^2$ different from each other and different from the $P_i$. Show that there is a cubic curve passing through all ten points $P_1, \ldots, P_8, P^{\prime}, P^{\prime \prime}$. (Hint. Show that there exist constants $\lambda_1, \lambda_2, \lambda_3$ such that $F=\lambda_1 F_1+\lambda_2 F_2+\lambda_3 F_3$ is not identically zero and such that the curve $F=0$ does the job.)
- (iii) Show that no four of the eight points $P_i$ are collinear, and no seven of them lie on a conic. (Hint. Use the fact that $C_1$ and $C_2$ have no common component.) (iv) Use the previous exercise to observe that there is a unique conic $Q$ going through any five of the eight points $P_1, \ldots, P_8$.
- (v) Show that no three of the eight points $P_i$ are collinear. (Hint. If three are on a line $L$, let $Q$ be the unique conic going through the other five, choose $P^{\prime}$ on $L$ and $P^{\prime \prime}$ not on $L$. Then use (ii) to get a cubic which has $L$ as a component, so is of the form $C=L \cup Q^{\prime}$ for some conic $Q^{\prime}$. This contradicts the fact that $Q$ is unique.)
- (vi) To get the final contradiction, let $Q$ be the conic through the five points $P_1, P_2, \ldots, P_5$. By (iii), at least one (in fact two) of the remaining three points is not on $Q$. Call it $P_6$, and let $L$ be the line joining $P_7$ to $P_8$. Choose $P^{\prime}$ and $P^{\prime \prime}$ on $L$ so that again the cubic $C$ through the ten points has $L$ as a component. Show that this gives a contradiction.
I think I have successfully followed all the steps, but I would like to see what a nicely written proof would look like to check my work. Does anyone know a paper where this proof has been presented in this method? I couldn't find anything in my searches.