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$$\Large\textbf{Problem}$$ Let $E:y^2 = x^3 + 1$ be an elliptic curve. For each prime $5 \leq p \leq 13$, describe the group $E(\mathbb{F}_p)$, the Mordell-Weil group.

$$\Large\textbf{Attempts and Ideas}$$ I am having lots of thoughts about this problem, so I jot them down.

Since $E(\mathbb{F}_p)$ is a nonempty group, the identity $\infty$ (this is how my professor defines the identity in Mordell-Weil group) must be in $E(\mathbb{F}_p)$.

For any prime $p$ such that $5 \leq p \leq 13$, we see that there is a point of order $2$ in the group $E(\mathbb{F}_p)$ since there exists $x \in \mathbb{F}_p$ such that $0 = x^3 + 1$ holds.

It remains to determine the points of finite order $n \geq 2$. If $E:y^2 = x^3 + ax^2 + bx + c$, then

$$\mathrm{Discr}_E = -4a^3b + a^2b^2 + 18ab - 4b^3 - 27c^2$$

Since $a = b = 0$ and $c = 1$, $\mathrm{Discr}_E = -27$. The strong form of Nagell-Lutz Theorem tells us that if $\beta^2 | -27$ and $(\alpha, \beta) \in E(\mathbb{F}_p)$, then $\beta \in \{\pm 1, \pm 3\}$. Write

$$\beta^2 = \alpha^3 + 1$$

We need to check if there are points of finite order. If $\beta = \pm1$, then $ \alpha$ does not exist since it does not belong to $\mathbb{F}_p$ for any prime $5 \leq p \leq 13$ (at this point $\alpha = 0 \not\in \mathbb{F}_p$). If $\beta = \pm 3$, then $\alpha^3 = 8$, which implies $\alpha = 2$.

Overall, $E(\mathbb{F}_p)$ consists of $\infty$, the point of order $2$ and two points of order $4$ (I believe that this is isomorphic to $\mathbb{Z}_4$).

$$\Large\textbf{Notes}$$

This problem is similar to "The Group of points of Elliptic curve $y^2 = x^3 + 1$ over $\mathbb{F}_5$", but different since the given problem asks me to describe the group for prime $5 \leq p \leq 13$ (hopefully not a duplicate!).

Other than that, leave comments/answers/thoughts about the details I posted here.

Community
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NasuSama
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    I suppose the first question is whether you need to find all the points by bruteforcing $\mathbb F_p\times \mathbb F_p$. If not, you will need to know that you have indeed found all of them. If you do not have access to Hasse's theorem, this seems quite hard to do. If you do bruteforce, at that point you are basically reduced to a group theory problem and it seems natural to just continue with group theoretic computations. You might want to also note that if $(x,y)\in E(\mathbb F_p)$ then $(x, p-y)\in E(\mathbb F_p)$. – Yong Hao Ng Mar 19 '14 at 10:34

1 Answers1

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Take for instance $p=5$. The curve $E: y^2=x^3+1$ has good reduction at $5$, so $E/\mathbb{F}_5$ is an elliptic curve, in particular, $E(\mathbb{F}_p)$ is a finite abelian group. You can easily find all the points of $E$ in $\mathbb{P}^2(\mathbb{F}_5)$, and verify that there are $6$ points (counting the point at infinity). Since $E(\mathbb{F}_p)$ is finite abelian of order $6$, it must be isomorphic to $\mathbb{Z}/6\mathbb{Z}$. Indeed, you can also verify that $P=(2,3)$ is a generator, with $$2P=(0,1),\ 3P=(4,0),\ 4P=(0,4),\ 5P=(2,2),$$ and $6P=\infty$. You can work out the structure of $E(\mathbb{F}_p)$, for any $p>3$ in a similar way. For example, you can verify that $E(\mathbb{F}_7)\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/6\mathbb{Z}$, or that $E(\mathbb{F}_{17})\cong \mathbb{Z}/18\mathbb{Z}$.

Álvaro Lozano-Robledo
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