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I am having trouble with exercise 5.35 in Algebraic Curves by Fulton, in the case where $C$ is reducible without multiple components.

For context, if $C$ is an irreducible cubic over some algebraically closed field and $C^\circ$ denotes the simple points on $C$, then we can endow $C^\circ$ with the structure of group in the following way: For $P,Q\in C^\circ$, define $P\ast Q$ to be the third point of intersection of $\overline{PQ}$ with $C$ (When $P=Q$, $\overline{PQ}$ is the tangent line of $C$ at $P$). Fix $O\in C^\circ$. For $P,Q\in C^\circ$, we define $$P\oplus Q := O\ast(P\ast Q).$$ To Check that $(C^\circ,\oplus)$ is a group, one checks that $\ast$ is well-defined and uses the same argument as in the proof of Proposition 4 on the preceding page.

My question: I don't really understand how one can do something similar when $C$ has a line $L$ as a component. If $P,Q\in L\cap C^\circ$, how should one go about defining $P\oplus Q$? There isn't a unique third point of intersection of $\overline{PQ}=L$ with $C$.

LeterPGOD
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  • The usual group on the unit circle arises in this way, $z(x^2+y^2-z^2)=0,$ the third intersection point then is the direction. Parallel lines express the addition of angles. Choosing $O=(1:0:1),$ and looking in the affine $z=1.$ – Jan-Magnus Økland Apr 16 '25 at 09:43
  • @jan-MagnusØkland right, then I would get a group structure on the affine part of this curve, i.e. on the curve $x^2+y^2-1$. What the exercise in question is suggesting is that the same construction that turns a non-singular cubic into group, works for the simple points of a reducible cubic without multiple components. The construction clearly extends to work on the simple points of an irreducible cubic, but I don't see it working for the reducible case. For instance how do I define what it should mean to apply the group operation to $[1:0:0]$ and $[0:1:0]$ in the example you give? – LeterPGOD Apr 16 '25 at 14:34
  • The singularities are the circular points at infinity, and those are excluded. The two points you mention are the directions of the $x$ and $y$-axes which work fine. Remember all lines go through $O=(1:0:1)$ so in the direction of the $x$-axis you get $(1:0:1),(-1:0:1),(1:0:0)$ and in the direction of the $y$-axis you get $2O,(0:1:0)$ – Jan-Magnus Økland Apr 16 '25 at 16:31
  • Then the offset (parallel) lines give for real offsets with real or double solutions the angles, and for complex intersections, what? Has someone doen this example thoroughly? It'd be fun to work out in detail – Jan-Magnus Økland Apr 16 '25 at 17:17

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