Why is it necessarily true that given a smooth cubic curve $S_{3} \subseteq \mathbb{P}^{3}_{\mathbb{C}}$, containing a line $L$, ($L \subset S_{3}$) that: $$ S_{3} \cap \Pi = L \cup C$$ For a conic $C$ and plane $\Pi$ where $L \subset \Pi$?
This is stated in my lecture notes without any proof. As I understand it the following is true:
1.The statement implies that the intersection of a plane and a cubic must itself define a homogeneous polynomial of degree $d \le 3$.
2.Bezout's theorem should imply $C_{1} \cap C_{2} \le d_{1}d_{2}$. $(d_{1} = \text{degree} \; S_{3} ,d_{2}=\text{degree} = \; \Pi $)
3.Any line $L$ will have an uncountable number of points, and the union will not remove any of these.
How do we reconcile (2) and (3) - or, better, how can I actually prove this?
Thanks!
Both notation and the situation you are describing suggest that $S_3$ must be a surface. In that case everything you said makes perfect sense and the cubic curve (union of line and conic) would be intersection of the surface and the plane.
Otherwise, if you pick a plane containing the line (this is not a general plane!) then the intersection with the "remaining" degree 2 part or your curve wouldn't be a conic, but just two points (for a general plane). So the point of this paragraph is that it shouldn't be difficult to come up with a counterexample to your claim, and that actually most of the examples you would try to write down would fail.