Exercise 5.7.1 (Elements of order 2 in the group $E_f(\mathbb{R})$)

Question

Let $f(x,y) = y^2 - p(x)$, where $p(x)$ is a cubic polynomial with no repeated root. Take the point $O$ on $\mathscr{C}_f(\mathbb{R})$ to be the point $0:1:0$ at infinity. Show that $2A = O$ if and only if $A$ is of the form $A = (r,0)$, where $r$ is a root of $p(x)$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

\begin{proof} 
We know that $\mathscr{C}_f(\R)$ is a group for the $+$ law, and the point $O$ on $\mathscr{C}_f(\mathbb{R})$ to be the point $0:1:0$ at infinity.

Let $A(r,s)$ be a point of the curve, distinct of the point at infinity $O$. Then the coordinates of $-A$ are $(r,-s)$.
$$2A = O \iff A = -A\iff (r,s) = (-r,s) \iff s =0.$$
Moreover $(r,0)$ is on the curve $\mathscr{C}_f(\R)$ if and only if $0 = p(r)$.

If $A 
e O$,  $2A = O$ if and only if $A$ is of the form $A = (r,0)$, where $r$ is a root of $p(x)$.
\end{proof}

Note 1: The tangent at a point $P = (r,0)$ is a vertical line, which contains $O = (0:1:0)$. This explains anew why $P + P = O$.

Note 2: Since the roots of $p$ are simple roots, there  are  1 or 3 points $(r,0)$ on the curve.

In the first case, the group of elements such that $2A = 0$ (elements whose order divides $2$) is isomorphic to $C_2$.

In the second case, this group is  isomorphic to $C_2 	imes C_2$.

Exercise 5.7.1 (Elements of order 2 in the group $E_f(\mathbb{R})$)

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Exercise 5.7.1 (Elements of order 2 in the group $E_f(\mathbb{R})$)

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