Exercise 5.7.13 ($x(2P) \in \mathbb{Q}^2$ if $c = 0$)

Question

Show that the formula for $u_3$ in (5.56) can be rewritten as $u_3 = (u_1^2 - b w_1^4)^2 - 4c(3u_1 + aw_1^2) w_1^6$. Deduce that if the equation (5.50) has integral coefficients, and if $P$ is a rational point on $\mathscr{C}_f(\R)$, then the $x$-coordinate of $2P$ is the square of a rational number if $c = 0$.

Richard Ganaye · Accepted Answer

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\begin{proof} 
Let $P_1 = (x_1,y_1) = (u_1/w_1^2,v_1/w_1^3)$ be a point of $\mathscr{C}_f(\Q)$, and $2P_1 = (x_3, y_3)$. By the explicit formulas (5.53), we have
\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(a+2x_1)y_1^2}{4y_1^2}\
&=\frac{\left(3\frac{u_1^2}{w_1^4} + 2a\frac{u_1}{w_1^2} + bight)^2 - 4\left(a+2\frac{u_1}{w_1^2}ight)\frac{v_1^2}{w_1^6}}{4\frac{v_1^2}{w_1^6}}\
&= \frac{(3u_1^2 + 2a u_1 w_1^2 +bw_1^4)^2 - 4 (aw_1^2 + 2u_1)v_1^2}{(2v_1w_1)^2}.
\end{align*}
So we may write $x_3 = x(2P_1) = u_3/w_3^2$, where (as in formulas (5.56))
\begin{align*}
u_3  &= (3u_1^2 + 2a u_1 w_1^2 +bw_1^4)^2 - 4 (aw_1^2 + 2u_1)v_1^2,\
w_3 &= 2 v_1w_1
\end{align*}
(Note that here $u_3$ and $w_3$ may have common factors.)

Since $P_1 = (x_1,y_1) = (u_1/w_1^2,v_1/w_1^3) \in \mathscr{C}_f(\Q)$, the equation of $f$ gives
$$\frac{v_1^2}{w_1^6} = \frac{u_1^3}{w_1^6} + a \frac{u_1^2}{w_1^4} + b \frac{u_1}{w_1^2} + c,$$
that is
$$v_1^2 = u_1^3 + au_1^2 w_1^2 + b u_1 w_1^4 + c w_1^6.$$
Therefore
$$u_3 = (3u_1^2 + 2a u_1 w_1^2 +bw_1^4)^2 - 4 (aw_1^2 + 2u_1)(u_1^3 + au_1^2 w_1^2 + b u_1 w_1^4 + c w_1^6).$$
If we expand the second member, we obtain
$$u_3 = u_1^4 - 2b u_1^2 w_1^4 - 8c u_1 w_1^6 + (b^2 - 4ac) w_1^8$$
(same numerator as in Problem 12).
Therefore
\begin{align*}
u_3 &= u_1^4 - 2b u_1^2 w_1^4 + b^2 w_1^8  - 4c(2u_1 + aw_1^2) w_1^6\
&= (u_1^2 - b w_1^4)^2 - 4c(2u_1 + aw_1^2) w_1^6.
\end{align*}

If $c = 0$, we obtain
$$u_3 = (u_1^2 - b w_1^4)^2.$$
This shows that 
$$x(2P_1) = x_3 = \left( \frac{u_1^2 - b w_1^4}{2v_1w_1} ight)^2$$
is the square of a rational number.
\end{proof}

Exercise 5.7.13 ($x(2P) \in \mathbb{Q}^2$ if $c = 0$)

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Exercise 5.7.13 ($x(2P) \in \mathbb{Q}^2$ if $c = 0$)

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