Question: Given: two points, $p=(a_1:b_2:1),q=(a_2:b_2:1)\in \mathbb{C}^2 \subset \mathbb{CP}^2$ and a cubic (i.e. elliptic) plane curve $\mathcal{C}$ (for example: $\{(x:y:z): x^3 - y^2z + z^3=0 \}$), such that $p,q \in \mathcal{C}$. What is a (or the) rigorous method to find the third point of the intersection of $\ell(p,q)$ (the unique line passing between $p$ and $q$) with $\mathcal{C}$ when this third point $r$ lies on the line at infinity?
In other words, how can one rigorously find the formula for $r=\ell(p,q)\cap \mathcal{C}\setminus\{p,q\}$ when $r \in \{(x:y:0)\in \mathbb{CP}^2 \}$, the line at infinity?
I know several ad hoc/intuitive/non-rigorous ways for finding $r$ which produce the correct answer. However, my usual rigorous method for finding $r$ breaks down when $r$ is in the line at infinity.
Background/Context/Motivation:
Let's say I have the points $P_2=(0:1:1)$ and $P_4=(0:-1:1)$ in the complex projective plane, which both lie on the cubic curve $\mathcal{C}=V(x^3 -y^2z +z^3)=\{(x:y:z)\in \mathbb{CP}^2 : x^3 - y^2z +z^3=0 \}$.
(This is from exercises 2.3.5 and 2.3.6 in Algebraic Geometry: A Problem Solving Approach.)
Now I don't know the equation for the line between $P_2$ and $P_4$ explicitly, but I can parametrize it as $\ell=\{ tP_2 + (1-t)P_4\in \mathbb{CP}^2 :t\in \mathbb{C} \}$, which is is actually easier to work with (usually) because the parametrization makes it univariate instead of trivariate.
Anyway, I know by Bezout's Theorem or the Fundamental Theorem of Algebra, that $\ell$ and $\mathcal{C}$ must intersect at exactly three points (up to multiplicity). The problem is this: the univariate polynomial whose zeroes are supposed to be the three points of $\ell \cap \mathcal{C}$ is in this case quadratic (it is usually cubic as one would expect), and reduces to $t(t-1)=0$, i.e. not only is there one root too few, but the two roots which one does find are the two trivial roots which exist necessarily by the chosen parametrization for $\ell$ (i.e. when $t=0$, $P_4$, and when $t=1$, $P_2$). The whole point is to find the third point of $\ell \cap \mathcal{C}$ because this is the product of $P_2$ and $P_4$ under the chord-tangent composition law. For some reason this method does not work when the intersection of the parametrized line and the plane curve of interest lies on the line at infinity -- why does it not work in this case?
For the record, what I am doing is substituting $\ell$ into the equation for $\mathcal{C}$, because (ostensibly) one should have that $$\{tP_2 + (1-t)P_4\in \mathbb{CP}^2 :t\in \mathbb{C} \} \cap \{ (x:y:z)\in \mathbb{CP}^2 : x^3 - y^2z +z^3=0 \} \\ = \{ (x:y:z) : \left[(x:y:z)=tP_2 + (1-t)P_4\text{ for some }t \in \mathbb{C}\right]\ \land\ \left[x^3 -y^2z + z^3 =0 \right] \} \\ = \{ (0:2t-1:1) : (0)^3-(2t-1)^2(1) + (1)^3 =0 \} $$ However, this doesn't work, again because the polynomial in the last line is quadratic and not cubic.
Now it is clear to me for several reasons that the third point of intersection of $\ell$ with $\mathcal{C}$ is $(0:1:0)$
- Looking at the picture of the situation, it is self-evident that the third point of intersection cannot be in $\mathbb{C}^2$ and thus must lie on the line at infinity. Moreover, one can check that $(0:1:0)$ is the only point in $\mathcal{C}$ which is also in the line at infinity.
- The point in the line at infinity corresponding to $P_2=(0:1:1)$ is $(0:1:0)$ and the point in the line at infinity corresponding to $P_4=(0:-1:1)$ is $(0:-1:0)=(0:1:0)$.
- The line between $P_2$ and $P_4$ corresponds to the line $x=0$ in $\mathbb{C}^2$, which can only intersect the line at infinity at $(0:1:0)$.
- For points of $\ell$ such that $t\not=\frac{1}{2}$, one has that $(0:2t-1:1)=(0:1:\frac{1}{2t-1})$. Then $\underset{t \to \infty}{\lim} (0:1:\frac{1}{2t-1})=(0:1:0)$ and because intuitively lines in projective space "wrap around and meet at infinity" one should have that this limit is in $\ell$ and thus is our "third root".
Ideally, I would like to setup the situation so that I have a cubic polynomial and that $(0:1:0)$ actually corresponds to the third root of this polynomial. However, I do not know how to do this.
Before we delve into your question about parametrizing lines in projective space, let's try to find the (implicit) equation of the line through $P_2$ and $P_4$. Since the last coordinates of $P_2$ and $P_4$ are both nonzero, we can work in the affine open set where $Z \neq 0$ with coordinates $x = X/Z$ and $y = Y/Z$. In this open set, $P_2 = (0,1)$ and $P_4 = (0,-1)$, and it should be easy to see that the line between them is simply $x = 0$. Homogenizing, we find $x = X/Z = 0$, so this is the line $X = 0$ in projective space.
Another approach to determining the line: we know that every line in $\mathbb{P^2}$ is of the form $aX + bY + cZ = 0$ for some $a,b,c \in \mathbb{C}$ (not all zero). Plug in the coordinates of $P_2$ and $P_4$, and you should find that $b = c = 0$, just as above.
Substituting $X = 0$ into the equation for the curve $C: X^3 - Y^2 Z + Z^3$, we find $$ 0 = -Y^2Z + Z^3 = Z(Z^2 - Y^2) = Z(Z-Y)(Z+Y) \, . $$ This yields 3 solutions: $Z = 0$, $Z = Y$, and $Z = -Y$, corresponding to the points $(0:1:0), (0:1:1), (0:-1:1)$ (recall that $X = 0$), the last two of which are $P_2, P_4$, and the first of which is the third point of intersection.
Now to your questions about parametrization. As you discovered, the formula $tP_1 + (1-t)P_2$ that you learned in analytic geometry doesn't quite work in projective space! As you noted, it is clear that $(0:2t-1:1)$ cannot be correct because it does not intersect the line at infinity $Z=0$, but every two lines in $\mathbb{P}^2$ intersect by Bézout's Theorem. (Also, note that the polynomial $2t-1$ is not homogeneous.) However, we can salvage the correct answer by recalling that a parametrization of a curve $V$ is really a (dominant rational) map $\mathbb{P}^1 \to V$. The parametrization $(0:2t-1:1)$ doesn't look like a map from $\mathbb{P}^1$ but that's because we are working in the affine open set of $\mathbb{P}^1$ where $t = S/T$ and $T \neq 0$ (where $S,T$ are the coordinates of $\mathbb{P}^1$). Substituting $t = S/T$, we find $$ (0:2t-1:1) = (0:2(S/T)-1:1) = (0 : 2S - T : T) $$ which gives the full parametrization, including the point $(0:1:0)$ when $T = 0$. Try substituting this parametrization into the equation for $C$ and check that we get the same answer as above. (If you have trouble factoring the resulting two-variable polynomial, remember that you can always dehomogenize, factor, then rehomogenize.)
The usual formula for the line between two points $P_1, P_2$ in projective space is $$ \ell = \{SP_1 + TP_2 \mid [S:T] \in \mathbb{P}^1\} \, . $$ You can check that this formula gives a slightly different, but equivalent parametrization of the line between $P_2$ and $P_4$.