Do two lines passing through two (not necessarily distinct) points on an irred. cubic necessarily coincide? (Proof of Sec. 5.6, Prop. 3 in Fulton)

Question

Prop. 3 of Sec. 5.6 of Fulton's "Algebraic Curves" is:

Let $C$ be an irreducible cubic, $C'$, $C''$ cubics. Suppose $C' ⋅ C = ∑_{i=1}^9 P_i$, where the $P_i$ are simple (not necessarily distinct) points on $C$, and suppose $C'' ⋅ C = ∑_{i=1}^8 P_i + Q$. Then $Q = P_9$.

Proof. Let $L$ be a line through $P_9$ that doesn't pass through $Q$; $L ⋅ C = P_9 + R + S$. Then $LC'' ⋅ C = C' ⋅ C + Q + R + S$, so there is a line $L'$ such that $L' ⋅ C = Q + R + S$. But then $L' = L$ and so $P_9 = Q$.

I can't see the justification for concluding $L' = L$ from $L ⋅ C = P_9 + R + S$ and $L' ⋅ C = Q + R + S$. Was $L$ chosen such that $R$ and $S$ are distinct? If so, that itself needs justification. Even though it seems intuitive that it would be possible over $\mathbb{C}$, it seems less intuitive over arbitrary fields, even algebraically-closed ones.

Otherwise, if $R$ and $S$ coincide, then we don't necessarily know that $R = S$ is a simple point, and so it might be a multiple point, like the origin for (the projectivization of) $Y^2 = X^3 - X^2$.

It's possible that we're meant to assume that $C$ is non-singular, in which case we can use the statement that there is a unique line $L$ through any two (not necessarily distinct) points. But that seems unlikely, since the very next paragraph talks about the nonsingular case.

It seems likely that I'm missing something obvious that Fulton is leaving unstated. Anyone know what that is?

(Note that the dot notation here is the intersection cycle; for plane curves $F$, $G$:

$$ F ⋅ G = ∑_{P \in \mathbb{P}^2} I(P, F ∩ G)P, $$ which by Bezout's Theorem has degree equal to the product of the degrees of $F$ and $G$.

Also, this section is on Max Noether's Fundamental theorem, which is what justifies the existence of the line $L'$.)

Answer 1

I think you did find a gap here, but it can be remedied without extended argument. There's no need to assume $R\neq S$ or $C$ nonsingular - all you need is to pick $L$ which does not contain a singular point of the cubic. (Since an irreducible plane cubic has one or zero singular points, this is always doable: you're trying to pick a line through a point which avoids at most two other points; there's always at least three lines through a given point no matter the base field.) Once you have that $L$ does not contain a singular point, then you have that if $R,S$ are distinct, any line through $R$ and $S$ is uniquely determined, and if $R,S$ are the same, then the only line meeting $R=S$ in multiplicity at least two is the tangent line to the cubic at $R=S$, which must exist since $R=S$ is not a singular point of our cubic.