# Why is the line connecting the identity point and another point vertical in an elliptic curve?

Addition of 2 points on an elliptic curve is described as follows:

$$L$$ is the line between $$P$$ and $$Q$$ and $$R$$

$$L'$$ is the line between $$O$$ and $$P + Q$$ and $$R$$

The book describes the algebraic process of adding together 2 points on an elliptic curve.

First: It describes adding together $$P$$ and $$Q$$ to get $$R$$. It then says we need to connect $$O$$ and $$R$$ with a line, and where that line intersects $$E$$ will be the point $$P + Q$$. So far so good.

It then says the line connecting $$O$$ and $$R$$ is vertical and is easy to describe in projective coordinates as $$x = x_3z$$ where $$R$$ is $$(x_3, y_3)$$.

The line connecting $$O$$ and $$R$$ is $$L'$$ doesn't seem to be vertical. Clearly, in the picture it's slanted downwards.

Does anyone know what's going on?

• What is going on is this: (a) On Fig. 8.1 and 8.2, the point $\mathcal O$ is explicitly drawn, i.e. is not an infinite point. The line goes through it; (b) If, as it usually is, the elliptic curve is given as $y^2=x^3+px+q$, then the point $\mathcal O$ is the infinite point in the direction of the $y$-axis. Thus, to draw a line towards $\mathcal O$, draw a vertical line. Jan 14 at 18:14

For complete generality, let’s write the equation of any line in the Cartesian plane thus: $$ax+by+c=0$$, with $$a$$ and $$b$$ not both zero. Now, remember that points $$(x,y)$$ in the Cartesian plane correspond to points $$(x:y:1)$$ in the projective plane. The projective version of the line above is $$aX+bY+cZ=0$$.

The neutral point of the elliptic curve is at $$\Bbb O=(0:1:0)$$. For a line as above to contain $$\Bbb O$$, the necessary and sufficient condition is that $$b=0$$.

Going back to the original Cartesian line, we see that its equation must be $$ax+c=0$$: vertical.

The construction and diagrams you're quoting are in Section 8.1 of the book.

In Section 8.2, the author changes the definition of $$\mathcal O$$ to $$(0:1:0)$$, i.e. the point at infinity in the vertical direction. In this new context, $$L'$$ becomes a vertical line.

This group theoretic construction in 8.1 works for nonsingular cubic curves in general where $$\mathcal O$$ can be any point on the curve. Elliptic curves are a special case of these curves, and always pass through (0:1:0). This simplifies the construction somewhat, in that the final stage is a simple reflection across the x-axis, rather than the intersection of the curve and a line.

• I guess there are various definitions of “elliptic curve”. The one I know is “curve of genus one with at least one rational point”. According to this definition, you wouldn’t need Weierstrass form; in particular, $x^3+y^3=1$ would be a good elliptic curve. Jan 16 at 2:48