Abelian torsion group of rational points of an elliptic curve

Question

I want to find the abelian group of rational points $E(\mathbb{Q})_{\text{torsion}}$ of the elliptic curve $y^2=x^3+8$.

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

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.

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So, to find the points $P$ of finite order of the curve $y^2=x^3+8$ we do the following:

For $y=0 \Rightarrow x^3+8=0 \Rightarrow x=-2$, $$P=(-2, 0) \in E(\mathbb{Q})$$ of order $2$

Let $f(x)=x^3+8$, then $D(f)=27 \cdot 8^2=2^6 \cdot 3^3$.

$$y^2 \mid 2^6 \cdot 3^3$$

$$2^i, i=0, 2, 4, 6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3^j, j=0, 2$$

$$y^2 \mid D(f) \Rightarrow y^2=1, 3^2, 2^2, 2^2 \cdot 3^2, 2^4, 2^4 \cdot 3^2, 2^6, 2^6 \cdot 3^2$$

$$ \Rightarrow y=\pm 1 , \pm 3, \pm 2, \pm 2 \cdot 3, \pm 2^2 , \pm 2^2 \cdot 3 , \pm 2^3, \pm 2^3 \cdot 3 \\ \Rightarrow y=\pm 1 , \pm 3, \pm 2, \pm 6, \pm 4 , \pm 12 , \pm 8, \pm 24 $$

For $y=\pm 1 \Rightarrow x^3+7=0 \Rightarrow x \notin \mathbb{Z}$.

For $y=\pm 3 \Rightarrow x^3-1=0 \Rightarrow x=1$, so there are two points with integer coordinates, $P_1=(1, 3), P_2=(1, -3)$.

Does this mean from Lutz-Nagell theorem that these two points are of finite order??

For $y=\pm 2 \Rightarrow x^3+4=0 \Rightarrow x \notin \mathbb{Z}$.

For $y=\pm 6 \Rightarrow x^3-28=0 \Rightarrow x \notin \mathbb{Z}$.

For $y=\pm 4 \Rightarrow x^3-8=0 \Rightarrow x=2$, so there are two points with integer coordinates, $P_3=(2, 4), P_4=(2, -4)$.

Which is the order of these points??

For $y=\pm 12 \Rightarrow x^3-136=0 \Rightarrow x \notin \mathbb{Z}$.

For $y=\pm 8 \Rightarrow x^3-56=0 \Rightarrow x \notin \mathbb{Z}$.

For $y=\pm 24 \Rightarrow x^3-568=0 \Rightarrow x \notin \mathbb{Z}$.

Answer 1

For this curve the torsion points are $P = (-2, 0)$ of order $2$ and the point at infinity. The reason why your points $P_1, ..., P_4$ are not torsion points are due to the fact that the Nagell-Lutz Theorem only gives you the possible candidates for the torsion points. In other words, Nagell-Lutz is not a if and only if-theorem.

When checking the list of possible torsion points, then if a point $P$ has order greater than $12$ then that point has infinite order. So you need to check if $nP = \mathcal{O}$, for $2 \leq n \leq 12$. If not, then $P$ has infinite order. This is due to the celebrated theorem of Barry Mazur, see wikipedia.

Answer 2

I'll try to rephrase to hopefully clarify in some detail how to use the Nagell-Lutz theorem and Mazur's theorem to find the torsion group.

So we work with the points $P_i$ that we find from Nagell-Lutz. We add $P_i$ to itself atmost $12$ times, that is for each $P_i$ we calculate the set $\{nP_i \text{ | } 2 \leq n \leq 12\}$. If in this set we find the point at infinity $\mathcal{O}$, we know that $P_i$ has finite order, by definition of a point of finite order. If we don't find $\mathcal{O}$ the order of $P_i$ must be $> 12$ and we conclude that the order of $P_i$ is infinite by Mazur's theorem.

Also to answer your last question, if $2P$ has infinite order then $P$ also has infinite order, because the torsion group is a subgroup. Or we can prove this easily by proof of contradiction.

Assume $2P$ has infinite order and assume to the contrary that $P$ has finite order. This means that we can find $n$ such that $nP = \mathcal{O}$. Then we calculate $n(2P) = 2(nP) = 2\mathcal{O} = \mathcal{O}$, so $2P$ has finite order, which contradicts our assumption. So $P$ must have infinite order.