defining a group by using the points of a parabola

Question

I'm trying to define an algebraic additive group on the parabola $(x + 1)(x - 1)$ by using the well-known geometric addition.

Given two points $P, Q$ on the parabola, I derived the equation of line that insertects both points, then took a parallel line that passes through the parabola vertex (0, -1), which is to be the group identity, and intersected this line with the parabola to get point which is the addition of $P, Q$. This gave me the following formula:

$$P(x_1, y_1) + Q(x_2, y_2) = \left(\frac{(y_2 - y_1)}{(x_2 - x_1)}, \frac{(y_2 - y_1)^2}{(x_2 - x_1)^2} - 1\right).$$

For adding $P + P$, I took the tangent at $P$, took a parallel line to it that passes through the vertex (0, -1) and intersected it with the parabola, getting the formula $P + P = (2x, 4x^2 - 1)$.

Taking a prime $p$, I expected to have a group reducing all points modulo $p$, but something goes wrong. For example, the point (0, -1) = (0, 4) mod 5 is the identity. Trying to see if (1, 0) is a generator, I compute (1, 0) + (1, 0) = (2, 3). Is (2, 3) on the parabola? (2, 2^2 - 1). It is. Reduction modulo 5 here doesn't change the coordinates. How about (1, 0) + (2, 3)? It's (3, 8), but reducing it mod 5 yields (3, 3) which is not on the parabola. (Maybe it is reducing the parabola itself modulo 5.)

Ignoring this possible problem, I see that (0, -1) = (0, 4) is not an identity. For instance, (0,4) + (2, 3) = (4, 4).

How could I build such group properly? Thank you.

Answer 1

Notice that, in your formula for $P(x_1,y_1)+Q(x_2,y_2)$, the first component $\frac{y_2-y_1}{x_2-x_1}$ is just $x_2+x_1$, because, thanks to the equation $y=x^2-1$ of your parabola, the numerator $y_2-y_1$ equals ${x_2}^2-{x_1}^2=(x_2+x_1)(x_2-x_1)$.

So your addition operation for points on the parabola is just adding the first components (note that this also works when the two points are equal), and so your parabola group is just a copy of the additive group of real numbers.

The same thing happens modulo any prime. To add two points, add their first coordinates to get the first coordinate of the answer, and then square that and subtract $1$ to get the second coordinate.

Answer 2

Notice that parabolas are affinely invariant. Given a parabola $P$, fix a point $P_0$. Given two other points $P_1,P_2$ on $P$, then the unique point $P_3$ on $P$ such that line $P_0P_3$ is parallel to line $P_1P_2$ defines an operation on points of $P$. The midpoints of the line segments $P_0P_3$ and $P_1P_2$ determine a line parallel to the axis of symmetry of $P$. Project the points $P_0,P_1,P_2,P_3$ onto the directrix of $P$ using lines parallel to the axis of symmetry and denote them as $X_0,X_1,X_2,X_3$ and note that $X_0+X_3=X_1+X_2.$ Introduce a coordinate system on the directrix, where $X_0$ is the origin. Denote the coordinates of the four points as $x_0,x_1,x_2,x_3$ where $x_0=0$. We get $x_3=x_1+x_2.$

This addition operation is similar in some ways to using the "prallelogram" law in affine geometry to add points. That is, given four points $P_0,P_1,P_2,P_3$ that are vertices of a parallelogram, with $P_0P_3$ and $P_1P_2$ the two diagonals, then $P_0+P_3=P_1+P_2,$ except in this case, the diagonals bisect each other and thus, the two midpoints are the same point.

There are only a few simple geometric ways to "add points" and your operation is a new one to me.