Let $E : y^2 = x^3 + Ax + B$ be an elliptic curve over a field $k$ with characteristic not $2$ or $3$ (not necessarily algebraically closed), and let $x = x_0$ be a line that intersects $E$ in $(x_0, \pm y_0) \in E(k)$.
According to Lemma 10 of this paper, this is expressed algebraically as the following equation of ideals over $k[X, Y]$: $$ (X - x_0, Y - y_0) (X - x_0, Y + y_0) \equiv (X - x_0) \pmod{(Y^2 - (X^3 + AX + B))}, $$ or more directly $$ (X - x_0, Y - y_0) (X - x_0, Y + y_0) = (X - x_0, Y^2 - (X^3 + AX + B)). $$
The paper gives a purely algebraic proof of this statement, although I can't help but think there must be a way to derive this algebro-geometrically.
I'd imagine it'd go something like: $$ V((X - x_0, Y - y_0) (X - x_0, Y + y_0)) = \\{ (x_0, \pm y_0) \\} = V(X - x_0, Y^2 - (X^3 + AX + B)) $$ We'd like to conclude that $$ (X - x_0, Y - y_0) (X - x_0, Y + y_0) = (X - x_0, Y^2 - (X^3 + AX + B)). $$
I know that $V(I) = V(J)$ doesn't imply $I = J$ in general; the most common condition is that $I$ and $J$ are radical ideals, and $k$ has to be algebraically closed, and then we can use the Nullstellensatz directly.
It seems doable to show that $(X - x_0, Y - y_0) (X - x_0, Y + y_0)$ is radical, via $$ \sqrt{I J} = \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J} $$ but it seems less obvious that $(X - x_0, Y^2 - (X^3 + AX + B))$ is radical. (Edit: It now seems doable by showing that $k[X, Y]/(X - x_0, Y^2 - (X^3 + AX + B))$ is reduced.)
Also, even if we assume that $k$ is algebraically closed and use the Nullstellensatz, maybe we can extend this to non-algebraically closed subfields of $k$ given some conditions, e.g. if the points of intersection are in the subfield. (Edit: this answer seems to be what I need.)
Am I on the right track?