Let $f:L\rightarrow \mathbb P^1_{\mathbb C}$ be the line bundle associated to the invertible sheaf $\mathcal O_{\mathbb P^1}(2)$, $\phi=(X_0X_1)^3X_0\in H^0(\mathbb P^1, \mathcal O_{\mathbb P^1}(4))$ and $C\subset L$ defined by the equation $T^2=f^*\phi$ where $T$ is the tautological section of $f^*\mathcal O_{\mathbb P^1}(2)$.
The curve $C$ is a singular rational curve, a double cover of the projective line ramified at $P=[1:1]$ and $Q=[0:1]$. As $Q$ seen in $C$ is a smooth point of $C$, if I am not mistaken, using $\mathcal O_C(3Q)$, we can embed $C$ in $\mathbb P^2=\mathcal O_C(3Q)$ as a cuspidal cubic curve.
Then given two smooth distinct points $P,P'$ of $C$, in the embedding in $\mathbb P^2$, we can consider the line $\ell\subset \mathbb P^2$ determined by $(P,P')$ and associate to this pair a third point (residual in the intersection $C\cap \ell$ in $\mathbb P^2$) of $C$.
Are there some interpretations of this line and this operation (associating a third point) related directly to $f:L\rightarrow \mathbb P^1$ and the definition of $C\subset L$?

$\begingroup$ The smooth part of $C$ is a cover of $P^1p$ that is doubly ramified over $q$, where I write $p=f(P)$ and $q=f(Q)$. If you identify the double cover with $A^1$ so that $Q$ corresponds to $0$, then the group law is addition. (The identification is unique up to scaling, which does not affect addition.) $\endgroup$– t3sujiMar 16, 2017 at 14:01
1 Answer
The smooth points on a cuspidal cubic curve in $\mathbb P^2$ can be given the structure of an algebraic group, just as is done for smooth cubic curves. If the tangent directions of the cuspidal point are defined over your ground field $K$, then you get a copy of $\mathbb G_m$. If not, the tangent directions generate a quadratic extension $L/K$, and the group you get is the $L/K$ twist of $\mathbb G_m$. See for example, The Arithmetic of Elliptic Curves, Proposition III.2.5 for the split case. As usual, the group law is determined by the condition that three points add to 0 if and only if they are colinear.

2$\begingroup$ If the point is cuspidal, there is only one tangent direction, and one obtains a copy of ${\mathbb G}_a$. $\endgroup$ Mar 16, 2017 at 13:44

$\begingroup$ @MichaelStoll Oops, sorry. You're right. I misread the post as saying the curve was nodal. For cuspidal curves, the cusp is always defined over the ground field and, as you say, the group is $\mathbb G_a$. This is described in the same reference. $\endgroup$ Mar 16, 2017 at 14:23