I just have a very basic question. For two distinct points $P_1, P_2$ with non-zero components on an elliptic curve $C$ given by $y^2 = x^3 + Ax + B$, I'm trying to derive the addition formula using the fact that the constant term of a monic cubic is equal to the negative of the product of the roots. Here's what I have so far, but I'm either making a mistake or missing the necessary algebraic trick.
Let $P_i = (x_i, y_i)$ for $i \in \{1,2\}$. Then the line from $P_1$ to $P_2$ is given by $y = m(x - x_1) + y_1$, where $m = (y_2 - y_1)/(x_2 - x_1)$. Then $$ y^2 = x^3 + Ax + B \implies f(x) = x^3 + Ax + B - (m(x-x_1) + y_1)^2$$ We can ignore all of the non-constant terms, so rewriting, we need to show that: $ (-mx_1 + y_1)^2 - B $ is the product of the roots. $$ m^2x_1^2 - 2mx_1y_1 + y_1^2 - B = m^2x_1^2 - 2mx_1y_1 + x_1^3 + Ax_1 $$ $$ \frac{m^2x_1^2 - 2mx_1y_1 + x_1^3 + Ax_1}{x_1x_2} = \frac{m^2x_1 - 2my_1 + x_1^2 + A}{x_2}$$ $$ \frac{m^2(x_2 + \frac{(y_1 - y_2)}{m}) - 2mx_1y_1 + x_1^3 + Ax_1}{x_1x_2} = \frac{m^2x_2 - my_1 - my_2 + x_1^2 + A}{x_2} $$ $$ = m^2 + \frac{-m(y_1 + y_2) + x^2_1 + A}{x_2} $$
I don't see how $\frac{-m(y_1 + y_2) + x^2_1 + A}{x_2} = -x_1 - x_2$ or even how to eliminate the $A$ term, so I must be making a mistake. Any help would be appreciated.
I’m writing this after almost a glass of wine, hope I don’t get the rest of it on my face.
This is just not the strategy I know, and that is by comparison so much shorter than what you’ve written that I don’t have courage to read the $n+1$ comments.
The line joining the $P_i$ will be given by $\ell:Y=mX+n$ for $n$ a rational expression that need not concern us. Then to get the intersection of $\ell$ with the curve of course you look for the roots of $g(X)=X^3+AX+B-(mX+n)^2$, when you know that $(X-x_1)(X-x_2)$ divides this. When you divide, you get a quotient $X-(m^2-x_1-x_2)$. You don’t need to do the division out — you just need to make sure that the $X^2$-coefficients are all right.
There’s your desired formula. If you try to use the constant term of $g$, you will, in my setup, get bogged down with the formula for $n$, a mess. I really don’t recommend doing it this way unless you are interested in an intellectual exercise that is, in my mind, going the long way round Robin Hood’s barn.