# Abelian group of rational points of an elliptic curve

I want to find the abelian group of rational points $E(\mathbb{Q})_{\text{torsion}}$ of the elliptic curve $y^2=x^3-2$.

$$E(\mathbb{Q})_{\text{torsion}}=\{P \in E(\mathbb{Q}) | P \text{ of finite order }\}$$

From Lutz-Nagell theorem we have the following:

Let $E|_{\mathbb{Q}} , y^2=x^3+ax+b, a, b \in \mathbb{Z}$ and $P=(x, y) \in E(\mathbb{Q})$.

We suppose that $P$ is of finite order.

Then $x, y \in \mathbb{Z}$ and $y=0$ (that corresponds to the points of order $2$ ) or $y^2 \mid D(f)=4a^3+27b^2$.

We have also the following:

If $P(x, y) \in E(\mathbb{Q})$ and $x \notin \mathbb{Z}$ or $y \notin \mathbb{Z}$, then $P$ has infinite order.



So, to find the points $P$ of finite order of the curve $y^2=x^3-2$ we do the following:

For $y=0 \Rightarrow x^3-2=0 \Rightarrow x \notin \mathbb{Z}$ (Do we have to say that $x \notin \mathbb{Z}$ or $x \notin \mathbb{Q}$ ?? )

so there is no rational point of order $2$.

Let $f(x)=x^3-2$, then $D(f)=27 \cdot 4=108$.

$$y^2 \mid D(f) \Rightarrow y^2 \mid 2^2 \cdot 3^3 \Rightarrow y^2 \mid 2^2 \cdot 3^2 \Rightarrow y \mid 6 \Rightarrow y=\pm 1, y =\pm 2, y=\pm 3 , y=\pm 6$$

For $y=\pm 1 \Rightarrow x^3-3=0 \Rightarrow x \notin \mathbb{Z}$.

For $y=\pm 2 \Rightarrow x^3-6=0 \Rightarrow x \notin \mathbb{Z}$.

For $y=\pm 3 \Rightarrow x^3-11=0 \Rightarrow x \notin \mathbb{Z}$.

For $y=\pm 6 \Rightarrow x^3-38=0 \Rightarrow x \notin \mathbb{Z}$.

So there are no rational points of the elliptic curve with finite order.

Is this correct??

• Is the point at infinity $O=[0, 1, 0]$ a point of the group $E(\mathbb{Q})_{\text{torsion}}$ ?? – Mary Star Jan 26 '15 at 18:48
• Yes, the torsion group, the subgroup of points of finite order, contains only the point at infinity. However, there are point of infinite order, namely $P = (3, 5)$ and the infinitely many multiples of $P$. – Jesper Petersen Jan 26 '15 at 18:56