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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!

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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.

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  • $\begingroup$ 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. $\endgroup$ – Shreya Apr 14 at 16:18
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    $\begingroup$ 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. $\endgroup$ – Alan Muniz Apr 14 at 16:26
  • $\begingroup$ 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? $\endgroup$ – Shreya Apr 14 at 20:19
  • $\begingroup$ 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? $\endgroup$ – Shreya Apr 14 at 20:34
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    $\begingroup$ 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. $\endgroup$ – Alan Muniz Apr 14 at 20:40

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