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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.