Exercise 5.6.6 (Rational points on the curve $y^2 = x^3 - 3x - 2$)

Question

Show that the curve $y^2 = x^3 - 3x - 2$ has an isolated double point. Use this double point to parameterize all rational points on the curve.

Richard Ganaye · Accepted Answer

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

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

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

\begin{proof} 
Put $f(x,y) = y^2 - x^3 + 3x + 2$. Then
$$\frac{\partial f}{\partial x}(x,y) = -3(x^2 - 1),\qquad \frac{\partial f}{\partial x}(x,y)  = 2y,$$
thus the only solution of $f(x,y) = \frac{\partial f}{\partial x}(x,y)  = \frac{\partial f}{\partial x}(x,y) = 0$ is $(x,y) = (-1,0)$.

The parametrization $y = t(x+1)$ starting from this double point gives 
\begin{align*}
(x,y) \in \mathscr{C}_f(\Q)  
&\iff (x,y) = (-1,0) 	ext{ or } \left( x 
e 0 	ext{ and }  \exists t \in \Q,\ 
\left\{
\begin{array}{l}
0 = t^2(x+1)^2 - x^3 + 3x + 2, \
y = t(x+1),
\end{array}
ight.
ight)
\
&\iff (x,y) = (-1,0) 	ext{ or } \left( x 
e 0 	ext{ and }  \exists t \in \Q,\ 
\left\{
\begin{array}{l}
0 = (x+1)^2(t^2-x+2), \
y = t(x+1),
\end{array}
ight.
ight)
\
&\iff (x,y) = (-1,0) 	ext{ or }  \exists t \in \Q, \ 
\left\{
\begin{array}{l}
x = t^2 + 2,\
y = t^3 + 3t.
\end{array}
ight.
\end{align*}
The rational points on the curve $y^2 = x^3 + 2x^2$ are $(-1,0)$ with the points $(t^2 + 2, t^3 +3t)$, where $t$ takes any rational value.
\end{proof}

Exercise 5.6.6 (Rational points on the curve $y^2 = x^3 - 3x - 2$)

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Exercise 5.6.6 (Rational points on the curve $y^2 = x^3 - 3x - 2$)

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