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

Question

I'm reading through Elliptic Tales.

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?

Answer 1

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.

Answer 2

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.