Three points on a line after homomorphism (elementary algebra)

Question

While going through the proof of Mordell's theorem on elliptic curves, I came across a certain homomorphism, and the problem is showing that this is indeed a homomorphism. I assure anyone reading this that NO elliptic curve knowledge is required since this is only an elementary algebra question. Let me introduce the map.

Background (Begin)

Let $E$ and $\overline{E}$ be elliptic curves given by the equations \begin{align*} E:y^2=x^3+ax^2+bx\quad\text{and}\quad\overline{E}:y^2=x^3+\overline{a}x^2+\overline{b}x,\quad a,b\in\mathbb{Q} \end{align*} where $\overline{a}=-2a$ and $\overline{b}=a^2-4b$. We define $T=(0,0)\in E$.

Then there is a homomorphism $\varphi:E\to\overline{E}$ defined by \begin{align*} \varphi(P)= \begin{cases} \left(\frac{y^2}{x^2},\frac{y(x^2-b)}{x^2}\right),\quad&\text{if }P=(x,y)\neq\mathscr{O},T\\ \overline{\mathscr{O}},&\text{if }P=\mathscr{O},T \end{cases} \end{align*} where $\ker(\varphi)=\{\mathscr{O},T\}$.

I already showed that the map is well-defined. No problem not knowing what $\mathscr{O}$ is since I already showed that \begin{align*} \varphi(P+Q)=\varphi(P)+\varphi(Q),\quad\forall P,Q\in\{\mathscr{O},T\}\text{ and for all }P\in E\text{ and all }Q\in\{\mathscr{O},T\}. \end{align*} So nothing with $T$ nor $\mathscr{O}$ is to be computed. Now it remains for me to show that \begin{align*} \varphi(P+Q)=\varphi(P)+\varphi(Q),\quad\forall P,Q\in E-\{\mathscr{O},T\} \end{align*} and I'm going to show this, knowing that $\varphi(-P)=-\varphi(P)$ for all $P\in E$ (without knowledge of elliptic curves, we can assume), by taking three points $P,Q,R\in E$ such that $P+Q+R=\mathscr{O}$ and show that \begin{align*} \varphi(P+Q)=\varphi(-R)=-\varphi(R)=\varphi(P)+\varphi(Q) \end{align*} which means exactly the same as showing that for three collinear points $P,Q,R$, that then $\varphi(P),\varphi(Q),\varphi(R)$ are also collinear.

Background (End)

Let's start. Let $y=\beta x+\gamma$ be the line on which $P=(x_p,y_p),Q=(x_q,y_q),R=(x_r,y_r)$ lie where $\beta=\frac{y_p-y_q}{x_p-x_q}$, $\gamma=y_p-\beta x_p$ and additionally we know that $R=(\beta^2-(a+x_p+x_q),\beta x_r+\gamma)$ by earlier computations I did for myself and I know they're true. Know I compute the points after applying $\varphi$: \begin{align*} P'=(X_P,Y_P)=\varphi(x_p,y_p)\\ Q'=(X_Q,Y_Q)=\varphi(x_q,y_q)\\ R'=(X_R,Y_R)=\varphi(x_r,y_r) \end{align*} Let $y=\delta x+\eta$ be the line on which both $P'$ and $Q'$ lie and thus $\delta=\frac{Y_P-Y_Q}{X_P-X_Q}$ and $\eta=Y_P-\delta X_P$. If I know compute $$Y_R-(\delta X_R+\eta)$$ with Mathematica, I don't get zero but it should be that $R'$ lies on the line $y=\delta x+\eta$ because it's in my book and John Tate said so... I even gave Mathematica extra assumptions on the points $P,Q,R,P',Q',R'$, the equations for the elliptic curves $E$. (Note that $P',Q',R'$ lie on $\overline{E}$ because the map is well-defined.) This should be easy but somehow it's also not since I don't get a proper result. Any help or suggestion is appreciated!

Answer 1

The following Mathematica code easily solves your problem in two ways.

(1) Using slope intercept given two points to find equation of line.

(2) Using the determinant of a $3\times 3$ matrix equaling zero.

ClearAll[a, b, phi, testE, testE2, rule, xp, yp, xq, yq, xr, yr];
a2 = -2*a; b2 = a^2 - 4*b;
phi[{x_, y_}] := {y^2, y (x^2 - b)}/x^2 // Simplify;
testE[{x_, y_}] := -y^2 + x^3 + a*x^2 + b*x;
testE2[{x_, y_}] := -y^2 + x^3 + a2*x^2 + b2*x;

rules = {yp -> Sqrt[xp^3 + a*xp^2 + b*xp],
        yq -> Sqrt[xq^3 + a*xq^2 + b*xq]};
(*" compute point R "*)
be = (yp - yq)/(xp - xq); ga = yp - be*xp;
xr = be^2 - (a + xp + xq); yr = be*xr + ga;
P = {xp, yp}; Q = {xq, yq}; R = {xr, yr};
M = {Ph = {xp, yp, 1}, Qh = {xq, yq, 1}, Rh = {xr, yr, 1}};

Print["P,Q,R collinear via Det[] is ", 0 == Det[M] /. rules // Simplify];
Print["P,Q,R on curve E is ", {0, 0, 0} == testE /@ {P, Q, R} /. rules // Simplify];

(*" collinearity coefficients for P,Q,R "*)
Sp = (xp - xq)^2*(a + xp + 2*xq) - (yp - yq)^2;
Sq = (xp - xq)^2*(a + 2*xp + xq) - (yp - yq)^2;
Sr = (xp - xq)^3;

Print["P,Q,R collinear via Sp,Sq,Sr is ", {0, 0, 0} == 
      Sp*Ph - Sq*Qh + Sr*Rh // Simplify];

(*" get P2,Q2,R2 via homomorphism phi[] "*)
P2 = phi[P]; Q2 = phi[Q]; R2 = phi[R];
Print["P2,Q2,R2 on curve E2 is  ", {0, 0, 0} == testE2 /@
  {P2, Q2, R2} /. rules // Simplify]  
(*" get coordinate for P2,Q2,R2 "*)
{Xp, Yp} = P2; {Xq, Yq} = Q2; {Xr, Yr} = R2;
M2 = {{Xp, Yp, 1}, {Xq, Yq, 1}, {Xr, Yr, 1}};
Print["P2,Q2,R2 collinear via Det[] is ", 0 == Det[M2] /. rules // Simplify];

(*" get slope and intercept "*)
de = (Yp - Yq)/(Xp - Xq) // Factor;
et = Yp - de*Xp // Factor;
Print["R2 is on line y=de*x+et: ", Yr == de*Xr + et /. rules // Simplify];