I am currently studying Bézout's theorem for projective plane curves and its applications. Amid this study, I came across the very famous Pascal's and Pappus's theorems. I was able to prove Pascal's theorem; as for Pappus's theorem, most of the authors I've read/watched state that this is just a particular case of Pascal's theorem, but I can't understand how.
The formulation of Pascal's theorem (according to this class notes from Andreas Gathmann) is:
Pascal's theorem. Let $F$ be an irreducible conic over an algebraically closed field. Pick six distinct points $P_1,\dots,P_6$ on $F$ (that can be thought of as the vertices of a hexagon inscribed in $F$). Then the three intersection points of the opposite edges of the hexagon (i.e., $P = \overline{P_1P_2} \cap \overline{P_4P_5}$, $Q = \overline{P_2P_3} \cap \overline{P_5P_6}$ and $R = \overline{P_3P_4} \cap \overline{P_6P_1}$) are colinear (all of this lie on a line).
Notice that the author restricts this to irreducible conics and this fact is indeed used in the proof (every other proof that I found using Bézout's theorem also uses this fact, even though some don't mention it (?)).
As for Pappus's theorem, it is formulated as follows:
Pappus's theorem. Let $A,B,C$ be three points on a straight line and let $X,Y,Z$ be three points on another line. If the lines $\overline{AY},\overline{BZ},\overline{CX}$ intersect the lines $\overline{BX},\overline{CY},\overline{AZ}$ respectively, then the three points of the intersection are colinear.
To prove this theorem, I was watching this really cool video from Zvi Rosen. Here, he says that Pappus's theorem is just a particular case of Pascal's theorem because obviously the union of two lines form a conic and then one can apply Pascal's theorem. Note that in this video Rosen stated Pascal's theorem without the word irreducible so it made sense to him to apply it directly, but in the proof of Pascal's theorem, he used the fact that the conic $F$ is irreducible. In fact, his proof follows exactly the same logic as the class notes linked above, from Gathmann.
My concern is that although the union of two lines forms a conic, this conic is reducible and thus one can't apply Pascal's theorem so directly (at least, in my point of view).
So, am I missing something here or we really can't apply the Pascal Theorem to prove the Pappus's theorem? If we can't apply it, how would one prove Pappus's Theorem using Bézout's theorem?
Thanks for any help in advance.
With a little more effort applied in the proof one can indeed recover Pappus's theorem as a special case of Pascal's theorem. Run the proof the same way as in Gathmann's note to start: let $A,B,C$ be on one line $\ell_1$, let $X,Y,Z$ be the other line $\ell_2$, and consider the two reducible cubics $G_1=\overline{AY}+\overline{BZ}+\overline{CX}$ and $G_2=\overline{AZ}+\overline{BX}+\overline{CY}$. For any $\theta$ on our two lines distinct from $A,B,C,X,Y,Z$ there are $\lambda_1,\lambda_2$ so that $G=\lambda_1G_1+\lambda_2G_2$ vanishes on $\theta$, and hence $F$ and $G$ meet in seven points, $A,B,C,X,Y,Z,\theta$. By Bezout, this implies that $F$ and $G$ share a common component, or at least one of the equations of $\ell_1$ or $\ell_2$ divides the equation for $G$. In fact, both do, so that $G=F\cdot L$ as before.
To see this, we examine what divisibility means. Any line can be written as the zero locus of a form $x+py+qz$ for $p,q\in k$ up to permutation of coordinates, so by the division algorithm we can write $\lambda_1G_2+\lambda_2G_2 = (x+py+qz)\cdot Q + R$, where $R=\sum_{i,j} \rho_{ij}(\lambda_1,\lambda_2)y^iz^j$ for $\rho_{ij}$ homogeneous polynomials in $\lambda_1,\lambda_2$. Then the statement that $x+py+qz$ divides $\lambda_1G_2+\lambda_2G_2$ is that $\rho_{ij}(\lambda_1,\lambda_2)=0$ for all $i,j$. But we can demonstrate that this occurs for a dense set of choices of the $\lambda_1,\lambda_2$ by choosing infinitely many different $\theta$ on our lines, which forces all the $\rho_{ij}$ to be identically zero. Hence the equations of both $\ell_1$ and $\ell_2$ divide $G$, and $G=F\cdot L$ as before.
Another perspective is to view both Pascal's and Pappus's theorems as consequences of another result:
The cubics are each of the two collections of intersecting lines as well as the conic union the line between two of the points of intersection. This is a bit more modern take on things and (to my taste) unifies them nicely as consequences of studying base points of linear systems.