Exercise 5.7.12 (Point of infinite order)

Question

Suppose that the elliptic curve $\mathscr{C}_f(\mathbb{R})$ is given by $y^2 = x^3 + ax^2 + bx +c$, and that the coefficients $a,b,c$ are integers. Let $P_1 = (u_1/w_1^2, v_1/w_1^3)$ be a rational point on this curve, and write $2P_1 = (u_3/w_3^2, v_3/w_3^3)$ with g.c.d.$(u_1,w_1) = 1$. Show that if $b$ is even and $u_1$ is odd, then $u_3$ is odd, and the power of $2$ in $w^3$ is greater than the power of $2$ in $w_1$. Deduce that the points $2^k P_1$ are all distinct, and hence that $P_1$ has infinite order. In particular, show that the point $(1,3)$ on the curve $y^2 = x^3 + 6x^2 + 2x$ has infinite order.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

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

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

\begin{proof} 
Now let $P_1 = (x_1,y_1)$ be any point on $\mathscr{C}_f(\Q)$, and suppose that $y_1 
e 0$, so $2P_1 = (x_3,y_3) 
e O$, where $O = (0:1:0)$ is the point at infinity. By Theorem 5.25, we may write 
\begin{align*}
x_1  = \frac{u_1}{w_1^2},\  y_1 = \frac{v_1}{w_1^3},\qquad  u_1 \wedge w_1 = 1,\ v_1 \wedge w_1 = 1,\
x_3 = \frac{u_3}{w_3^2},\  y_3 = \frac{v_3}{w_3^3}, \qquad  u_3 \wedge w_3 = 1,\ v_3 \wedge w_3 = 1.
\end{align*}
By the explicit formulas (5.53), we have (knowing that $y_1 
e 0$)
\begin{align*}
x_3 &= \left(\frac{3x_1^2 + 2a x_1 + b}{2y_1 }ight)^2 -a - 2x_1,\
&=\frac{(3x_1^2 + 2ax_1 +b)^2 -4 (x_1^3 + a x_1^2 + b x_1+ c)(a + 2x_1)}{4(x_1^3 + a x_1^2 + b x_1+ c)}\
&=\frac{x_1^4 - 2 b x_1^2 -8cx_1+b^2-4ac}{4x_1^3 + 4a x_1^2 + 4 b x_1 + 4c}.
\end{align*}
Substituting the values of $x_1, x_3$, we obtain
\begin{align*}
\frac{u_3}{w_3^2} &= \frac{\frac{u_1^4}{w_1^8} - 2 b \frac{u_1^2}{w_1^4} -8c\frac{u_1}{w_1^2}+b^2-4ac}{4 \frac{u_1^3}{w_1^6} + 4a \frac{u_1^2}{w_1^4} + 4 b \frac{u_1}{w_1^2}+ 4c}\
&= \frac{u_1^4 - 2b u_1^2 w_1^2 - 8c u_1 w_1^6 + (b^2 - 4ac) w_1^8}{4w_1^2(u_1^3 + a u_1^2 w_1^4 + b u_1 w_1^4 + c w_1^6)},
\end{align*}
so
\begin{align}
4w_1^2u_3(u_1^3 + a u_1^2 w_1^4 + b u_1 w_1^4 + c w_1^6)  = w_3^2[u_1^4 - 2b u_1^2 w_1^2 - 8c u_1 w_1^6 + (b^2 - 4ac) w_1^8].
\end{align}
Assume that $b$ is even and $u_1$ is odd. If we reduce equation (1) modulo $2$, we obtain
\begin{align*}
0 &\equiv w_3^2 (u_1^4 + b^2 w_1^8)\
&\equiv w_3^2\qquad  \pmod 2.
\end{align*}
Therefore $w_3$ is even, and since $u_3 \wedge w_3 = 1$, $u_3$ is odd.
\begin{itemize}
\item If $w_1$ is odd, then, since $w_2$ is even, $
u_2(w_2) >0 = 
u_2(w_1)$.
\item If $w_2$ is even, knowing that $u_1$ is odd, we obtain that $u_1^3 + a u_1^2 w_1^2 + b u_1 w_1^4 + c w_1^6$ and $u_1^4 - 2b u_1^2 w_1^2 - 8c u_1 w_1^6 + (b^2 - 4ac) w_1^8$ are odd. Therefore equation (2) gives
$$ 2 
u_2(w_3) = 2 
u_2(w_1) + 2,$$
so
$$ 
u_2(w_3) = 
u_2(w_1) + 1 > 
u_2(w_1).$$
\end{itemize}
In both cases, $
u_2(w_3) > 
u_2(w_1)$: the power of $2$ in $w_3$ is greater than the power of $2$ in $w_1$.

Since $x_1 = u_1/w_1^2$  and $x_3 = u_3/w_3^2$ are in lowest terms, this proves that $x_1 
e x_3$, hence $2 P_1 
e P_1$.

It is important to note that $y_3 
e 0$, because $y_3 = v_3 / w_3^3$, where $v_3 \wedge w_3 = 1$ and $w_3$ is even, therefore $v_3$ is odd, so $v_3 
e 0$ and $y_3 
e 0$. Therefore $2P_3$ is not the point $O$.

Hence we may continue with $P_3$ in place of $P_1$, and obtain in the same way that $
u_2(x(2P_3)) > 
u_2(x(P_2))$.

We obtain by induction that no point $2^k P_1$ is on the $y$-axis, and $
u_2(x(2P_k)) > 
u_2(x(P_k))$ for all $k$. The sequence $(
u_2(2^k P))_{k\in \N}$ is strictly increasing, so the points $2^k P_1$ are all distinct.

We conclude that $P_1$ has infinite order.

\bigskip
Example: Consider the  curve $\mathscr{C}: y^2 = x^3 + 6x +2 = x^3 + ax^2 +bx+c$. The discriminant of $ x^3 + 6x +2$ is $D = -4\cdot 6^3 - 27\cdot 2^2 = -972 
e 0$, thus $\mathscr{C}$ is an elliptic curve. $\mathscr{C}$  contains the point $P_1 = (x_1,y_1) = (1,3) = (1/1^2, 3/1^3) = (u_1/w_1^2 , v_1/w_1^3)$.  Then $b = 6$ is even and $u_1 = 1$ is odd (and $y_1 
e 0$). By Problem 12, $P_1$ has infinite order.
\end{proof}

\bigskip
With Sagemath:
\begin{verbatim}
sage: E = EllipticCurve([6,2]); E
Elliptic Curve defined by y^2 = x^3 + 6*x + 2 over Rational Field
sage: plot(E)
Launched png viewer for Graphics object consisting of 1 graphics primitive
sage: P = E(1,3)
sage: 2 * P
(1/4 : -15/8 : 1)
sage: 4 * P
(889/400 : 41037/8000 : 1)
\end{verbatim}

Exercise 5.7.12 (Point of infinite order)

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Exercise 5.7.12 (Point of infinite order)

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