Let me consider the elliptic curve $C: y^2=x^3+2x$. Now I want to finde $P\in C(\Bbb{Q})$ s.t. $2P=\infty$.
I thought about writing $P=\left(\frac{a}{b}, \frac{c}{d}\right)$, then I can use the addition formula and I get $2P=\left(\left(\frac{3\frac{a^2}{b^2}+2}{\frac{2c}{d}}\right)^2-\frac{2a}{b}, -\frac{c}{d}+\left(\frac{3a^2}{b^2}+2\right)\cdot \frac{d}{2c}\left(\frac{a}{b}-\left(\frac{3a^2}{b^2}+2\right)^2\cdot \frac{d^2}{4c^2}\right)\right)$ Where I used the formula that if $y^2=x^3+ux+v$ and $P=(x,y)$ then $2P$ has $x$-coordinate $x'=\left(\frac{3x^2+u}{2y}\right)^2-2x$ and has $y$-coordinate $y'=-y+\left(\frac{3x^2+u}{2y}\right)(x-x')$. But now I don't see how to finish this exercise, if I remember correctly then $2P=(x',y')=\infty$ iff $y'=0$ but I'm not sure if this is correct and if yes what this helps me.
Cans maybe someone help me?
I’m new in the world of elliptic curves but I can try:
I think that the formula that you want to use is not valid for $2$-torsion point. Infact if a point $P$ is such that $2P=0$ then we have that $y=0$, as you say in your question. But now we have a condition on $2P$ that we can use in $y^2=x^3+2x$. So, for $2P$ we have to have $$0=x^3+2x$$ and the unique rational solution is for $x=0$. So you want the point $P$ such that $2P=(0,0)$. So if $P=(x,y)$, and since in our case $u=2$ we obtain $$\begin{cases} & 0=\Big( \frac{3x^2+2}{2y}\Big)^2 \\ & 0=-y+\Big(\frac{3x^2+2}{2y}\Big)x \end{cases} $$ By the first equation we obtain $\frac{3x^2+2}{2y}=0$ that we can substitute in the second and we obtain $y=0$ that is not a valid solution. So I think that the sum formula is valid for point that are not $2$-torsion point.
I think that if a point is a $2$-torsion point we have to use the tangent line. So $P+P=-(t\times C)$ where $t$ is the tangent to $C$ in $P$. So if $g(X,Y)=Y^2-X^3-AX-B$ and $P=(P_X,P_Y)$ then the tangent is $(\nabla g(P)\cdot ((x,y)-P))=3P^2_Xx+Ax-3P_X^3-AP_X+2P_Yy-2P_Y^2$.
We want the points that have vertical tangent. Now, if $P_Y=0$ we obtain $x=P_X$. So we arrive another time $0=x^3+2x$. So, I think that the unique $2$ torsion point on $\mathbb{Q}$ is $(0,0)$.
I repeat that I’m new in this theory and I might have been wrong.