# Proving Pascal's theorem from a corollary to Max Noether's Fundamental theorem

A result in my book Fulton's Algebraic Curves states that:

If $$F$$ and $$G$$ meet in $$\deg(F\deg(G)$$ distinct points, and $$H$$ passes through these points then there exists a curve $$B$$ such that $$B•F = H•F -G•F$$ where $$B•F= \sum_{P \in F \cap B} I(P, B\cap F) P$$ where $$I(P, F\cap B)$$ is the intersection number of projective plane curves $$F$$ and $$B$$.

Another result says, all the points of $$F\cap G$$ are simple points of $$F$$, and $$H•F \geq G•F$$ then it implies that there exists a curve $$B$$ such that $$B•F= H•F-G•F$$

I have already proved these results. Now, using them I have to show Pascal's theorem which says that if a hexagon is inscribed in an irreducible conic then the opposite sides meet in collinear points.

My attempt: Choose any three sides of hexagon, they form a conic $$C$$ and the three sides of it form a conic $$C'$$ and let $$Q$$ be this given irreducible conic.

Now, since $$Q$$ and $$C$$ can have no common components so by Bezout's theorem it follows that $$Q•C= P_1 +... +P_6$$, I want to use Bezout's theorem again on $$C$$ and $$C'$$ to have $$C•C' = Q_1 + ...Q_9$$

And then use one of the results above to have a curve $$B$$ for which $$B•F =$$ a sum of three points.

My questions- 1. I can see that my method works if six of the points among $$Q_1,...Q_9$$ are $$P_1,...P_6$$ but I don't know how to show that?

1. Also, how can I apply Bezout's theorem on $$C$$ and $$C'$$? Doesn't it require for them to have no components common? How can I have that here?

I'd appreciate any help in the form of hint/s or solutions. Thanks!

Let $$C$$ and $$C'$$ be a decomposition in opposite sides of the hexagon. Although they are singular cubics, they meet in in smooth points and six among them are vertices of the hexagon, by hypothesis. $$C\cdot C' = P_1 + \dots + P_9$$ and let's say that $$P_1, \dots, P_6$$ are the vertices of the hexagon.
Now we use the theorem with $$F=C$$, $$G=Q$$ and $$H=C'$$. Then there exists a curve $$B$$ such that $$B\cdot C = C\cdot C' - C\cdot Q = P_7 + P_8 + P_9$$ By Bezout's theorem, $$B$$ must have degree one since $$C$$ is a cubic. Hence $$B$$ is a line and $$P_7, P_8, P_9$$ are colinear.
• Just to be sure, we choose $C$ to be any three sides and $C'$ to be the other three sides, right? Also, I'm not exactly sure what you and the book mean by 'opposite sides', I guess it's clear in case of regular hexagon but otherwise...I don't know. – Shreya Apr 14 at 16:18
• No, you choose three sides that are not adjacent so the singular set of $C$ does not lie on the hexagon. If the hexagon has sides $S_1, \dots, S_6$ such that the vertices are $S_i \cap S_{i+1}$ (indices taken mod six) then we may choose, for instance $C$ to be the odd sides and $C'$ to be the even sides. – Alan Muniz Apr 14 at 16:26
• Sorry if it's a dumb question, but how does choosing non-adjacent three sides for $C$ ensure that the singular set of $C$ does not lie in the hexagon? – Shreya Apr 14 at 20:19
• Also, why did you write $C•C' =$ a sum of $9$ points, isn't it only guranteed when Bezout's theorem is applicable but in this case it need not be since $C$ and $C'$ can have a common component? – Shreya Apr 14 at 20:34
• Well, $C$ is the union of three lines and its singular set is composed by their pairwise intersections. So a singular point of C is either a vertex or does not lie on the hexagon. Choosing the non adjacent ones ensures that we are in the later. – Alan Muniz Apr 14 at 20:40