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??


Yes, it is correct that there are no rational points of finite order. And your application of the Nagell-Lutz Theorem seems valid too.

  • $\begingroup$ Is the point at infinity $O=[0, 1, 0]$ a point of the group $E(\mathbb{Q})_{\text{torsion}}$ ?? $\endgroup$ – Mary Star Jan 26 '15 at 18:48
  • $\begingroup$ Yes, that is the identity element in the group structure. And that is regardless of the specific curve you are looking at. $\endgroup$ – Jesper Petersen Jan 26 '15 at 18:49
  • $\begingroup$ So, the group contains only the point at infinity, right?? $\endgroup$ – Mary Star Jan 26 '15 at 18:51
  • $\begingroup$ 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 $P$. $\endgroup$ – Jesper Petersen Jan 26 '15 at 18:56
  • $\begingroup$ Ahaa... Ok!! Thank you very much!!! :-) $\endgroup$ – Mary Star Jan 26 '15 at 21:04

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