The group structure of elliptic curve over $\mathbb F_p$

Question

I want to find the group of the elliptic curve $y^2=x^3-x$ over $\mathbb F_p$ for all primes $p \le 29$. But I know only 1 fact about the structure of this group: $E(\mathbb F_p)=\mathbb Z/m \mathbb Z \times \mathbb Z/nm\mathbb Z$ for $\gcd (m,p)=1, p =1 \mod m$. It doesn't really help. Is there any method to find the group without huge calculation?

Answer 1

The elliptic curve $E:y^2=x^3-x$ has discriminant $\Delta=64$, so it only has bad reduction at $p=2$. For any other $p>2$, the curve has good reduction and $E/\mathbb{F}_p$ is an elliptic curve, in particular, $E(\mathbb{F}_p)$ is a finite abelian group.

Take for instance $p=5$. You can easily find all the points of $E$ in $\mathbb{P}^2(\mathbb{F}_5)$, and verify that there are $8$ points (counting the point at infinity). Since $E(\mathbb{F}_5)$ is finite abelian of order $8$, it must be isomorphic to $(\mathbb{Z}/2\mathbb{Z})^3$, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$, or $\mathbb{Z}/8\mathbb{Z}$. The first possibility is not really possible, because the $q$ torsion of an elliptic curve (for any prime $q$) is formed by the direct sum of, at most, two cyclic groups of order $q$ (i.e., $E[q]\subseteq \mathbb{Z}/q\mathbb{Z}\times \mathbb{Z}/q\mathbb{Z}$). Now it is a matter of identifying $E(\mathbb{F}_5)$ as $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$, or $\mathbb{Z}/8\mathbb{Z}$, and you can prove that in fact $E(\mathbb{F}_5)\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$, by finding $3$ distinct points of exact order $2$ (if the group was $\mathbb{Z}/8\mathbb{Z}$ there would be only one point of exact order $2$).

Similarly, when $p=11$, you can count $E(\mathbb{F}_{11})$ and see that there are $12$ points (don't forget about the point at infinity!). Now the possible isomorphism classes are $\mathbb{Z}/12\mathbb{Z}$, or $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/6\mathbb{Z}$, and you can prove that $E(\mathbb{F}_{11})\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/6\mathbb{Z}$ by finding $3$ distinct points of order $2$.

Answer 2

For variety, we can compute $E(\mathbf{F}_3)$ by using Hasse's theorem, which says it has between $4 - 2\sqrt{3}$ and $4 + 2\sqrt{3}$ points: so it's somewhere from $1$ to $7$ points.

Because the curve has three points of order $2$, $\mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z}$ must be a subgroup. And thus it must be the entire group, since the only possible group order divisible by $4$ is $4$ itself.

Answer 3

Note that the point at infinity, together with the points $(-1,0),(0,0),(1,0)$ form a subgroup of $ E(\Bbb Q)$ isomorphic to $(\Bbb Z/2\Bbb Z)^2$, you will find this subgroup in $E(\Bbb F_p)$ for any $p > 2$.

Moreover, your curve has complex multiplication (by $i : (x,y) \mapsto (-x,iy)$), so a lot can be said via the explicit class field theory for $\Bbb Q(i)$.

If $p \equiv 3 \pmod 4$, the number of points on $E(\Bbb F_p)$ is $p+1$. Since there is no prime factor in common between $p+1$ and $p-1$ except possibly $2$, and we know that $\Bbb (\Bbb Z/2\Bbb Z)^2$ is a subgroup, but not $(\Bbb Z/4\Bbb Z)^2$ (because $p-1 \equiv 2 \pmod 4$), we must have $E(\Bbb F_p) \approx (\Bbb Z/2\Bbb Z) \times (\Bbb Z/\frac{p+1}2\Bbb Z)$.

The points $(i,1-i),(i,-1+i),(-i,1+i),(-i,-1-i)$ with the previous four form a subgroup of order $8$ in $E(\Bbb Q(i))$ (isomorphic to $(\Bbb Z/2\Bbb Z) \times (\Bbb Z/4\Bbb Z)$), so this subgroup will always be present in $E(\Bbb F_p)$ with $p \equiv 1 \pmod 4$

If $p \equiv 1 \pmod 4$, then there is a unique way to write $p = (a+ib)(a-ib)$ with $a+ib \equiv 1 \pmod {2+2i}$. Then, $\# E(\Bbb F_p) = p+1-2a$, and $E(\Bbb F_p)$ contains a factor $(\Bbb Z/m\Bbb Z)^2$ if and only if $(a,b) \equiv (1,0) \pmod m$ (the Frobenius automorphism $Frob_p$ acts on $E(\overline{\Bbb F_p})$ a lot like multiplication by $a+ib$ does on the torus $\Bbb Q[i] / \Bbb Z[i]$).