Let $P_1 = (x_1,y_1)$ and $P_2 = (x_2,y_2)$ be points on the elliptic curve $$y^2 + a_1xy+a_3y = x^3+a_2x^2+a_4x+a_6.$$

(I'm suppressing the homogenizing variable $z$; perhaps this is a bad idea? Feel free to treat everything I say as if the above were a homogeneous polynomial in three variables.)

Suppose we're trying to find $P_1\oplus P_2$.

If I understand correctly, there's three cases:

  1. If $x_1 = x_2$ and $y_1+y_2+a_1x_2+a_3 = 0,$ then $P_1 \oplus P_2 = O$.
  2. If $x_1 = x_2$ and $y_1+y_2+a_1x_2+a_3 \neq 0,$ then there's a complicated formula.
  3. If $x_1 \neq x_2$, then there's a different formula.

I don't understand the need for (2). Geometrically, it seems to me that if $x_1=x_2$, then the line through $P_1$ and $P_2$ will be vertical, and hence $P_1 \oplus P_2$ will equal $O$. Am I mistaken about this?

  • $\begingroup$ The 3rd point on $E : f(x,y)=0$ is given by the 3rd root of the cubic $g(T) = f(x_1+(x_2-x_1)T,y_1+(y_2-y_1)T)$. In the degenerate case, perturb $x_2$ and use limits, or projective coordinates, as you wish. $\endgroup$ – reuns Sep 21 '17 at 13:06
  • $\begingroup$ what's the equation of the curve ? $\endgroup$ – mercio Sep 21 '17 at 13:30
  • $\begingroup$ The way I've usually seen the 3 cases stated is: (1) $P_1 \neq P_2$ and the line between them is not vertical (your case 3); (2) $P_1 \neq P_2$ but the line between them is vertical, so $P_1 \oplus P_2 = O$ (your case 1) and; (3) $P_1 = P_2$, where instead we take the tangent line at $P_1$, (presumably your case 2). $\endgroup$ – André 3000 Sep 21 '17 at 14:09
  • $\begingroup$ @mercio, I've edited with the equation. $\endgroup$ – goblin Sep 21 '17 at 14:14
  • $\begingroup$ It's possibly a typo between $a_1$ and $a_2$ in (1) and (2)'s conditions. Then it would match with @Quasicoherent's comment I think. $\endgroup$ – mercio Sep 21 '17 at 14:18

Something is a bit funny about your second case. Mercio got it right I think. Adding a bit more details.

When $x_1=x_2=A$ both $y_1$ and $y_2$ are solutions of the equation $$ y^2+(a_1A+a_3)y=A^3+a_2A^2+a_4A+a_6.\qquad(*) $$ By Vieta relations the sum of the solutions of $(*)$ is the negative of the coefficient of the linear term. If $P_1\neq P_2$, then those solutions are $y=y_1$ and $y=y_2$. In other words $$ y_1+y_2=-(a_1A+a_3)\Longleftrightarrow y_1+y_2+a_1x_2+a_3=0. $$

  • So if $P_1\neq P_2$ and $x_1=x_2$, the condition $y_1+y_2+a_1x_2+a_3=0$ HOLDS ALWAYS. In this $P_1\oplus P_2=O$.
  • In other words, $x_1=x_2,y_1+y_2+a_1x_2+a_3\neq0$ is just a complicated way of saying that $P_1=P_2$. In this case the point addition formula relies on using the tangent.
  • As mercio also pointed out we can have the equation $y_1+y_2+a_1x_2+a_3=0$ also when $P_1=P_2$, but then $y_1=y_2$ is a double root of $(*)$. This is the case of a vertical tangent, and we get $P_1\oplus P_1=O$.

Anyway, defining the case 2 with an inequation $y_1+y_2+a_1x_2+a_3\neq0$ feels a bit unnatural. It might make sense in a computer program, because that is the only case where you need to do arithmetic with a non-vertical tangent. For the purposes of visualizing what's going on in case 2 it's IMHO more natural to say that we are doubling the point $P_1$ here. More often than not we have a formula for adding two distinct points, and another formula for doubling a point. Both have as a subcase the possibility of a vertical line used in finding the third point of intersection (= the point at infinity). I guess the advantage of this method is to have three cases as opposed to four (when you count the vertical case as splitting the the two main cases in two).


The case (2) occurs when $P_1 = P_2$ and the tangent line there is not vertical (if it is vertical it is already covered by case (1)).

In this case you intersect the cubic with its tangent line at $P$ and do some computation to find the third point, hence a complicated formula.


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