2
$\begingroup$

In the book Rational Points on Elliptic Curves by Silverman/Tate one examines the elliptic curve $y^2 = x^3 + x + 1$ over $F_5$. One can then easily determine the group $$ C(F_5) = \lbrace \mathcal{O}, (0, \pm 1), (2, \pm 1), (3, \pm 1), (4, \pm 2)\rbrace $$ which is an abelian group of order 9. The book claims that $C(F_5)$ is either abelian of order 9 or two cyclic groups of order 3 which makes sense. However, i can't understand the reasoning behind the fact that it must be a abelian group of order 9. There is two points of order $3$ and the rest are of order $9$.

Can anyone assist me with this reasoning. Thanks!

$\endgroup$
2
$\begingroup$

It is known in general that the group $E(\mathbb{F}_q)$ for an elliptic curve $E$ over $\mathbb{F}_q$ is either cyclic or isomorphic to $\mathbb{Z}/k\mathbb{Z} \times \mathbb{Z}/\ell\mathbb{Z}$ for some $k,\ell$. For $y^2=x^3+x+1$ over $\mathbb{F}_5$ the group is isomorphic to $\mathbb{Z}/9\mathbb{Z}$, so it is cyclic. Indeed, $P=(0,1)$ is a generator of order $9$. Since $\langle P \rangle \subseteq E(\mathbb{F}_5)$, and both groups have $9$ elements, it follows that $\mathbb{Z}/9\mathbb{Z}\simeq \langle P \rangle\simeq E(\mathbb{F}_5)$.

$\endgroup$
  • $\begingroup$ I just don't understand why it is isomorphic to ℤ/9ℤ and not ℤ/kℤ×ℤ/ℓℤ. $\endgroup$ – user111435 Dec 13 '13 at 11:03
  • $\begingroup$ The second group $\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}$ of order $9$ is not cyclic, because it does not have an element of order $9$. But our group has one. The subgroup generated by $P$ is cyclic, has order $9$, hence must coincide with $\mathbb{Z}/9\mathbb{Z}$. $\endgroup$ – Dietrich Burde Dec 13 '13 at 12:17
  • $\begingroup$ Thanks. That makes perfect sense! $\endgroup$ – user111435 Dec 13 '13 at 12:57
  • $\begingroup$ Here we can actually rule out the possibility $\Bbb{Z}/3\Bbb{Z}\times\Bbb{Z}/3\Bbb{Z}$ "without looking" by applying two somewhat non-trivial facts. The torsion group of any elliptic curve is of the form $\Bbb{Z}/k\Bbb{Z}\times\Bbb{Z}/\ell\Bbb{Z}$, where $\ell\mid k$. This is because we can bound the number of torsion points of a given order over the algebraical closue using the degree of the division polynomial. Furthermore, here always $k\mid q-1$, if we are looking at torsion over $\Bbb{F}_q$. This follows from the existence of the Weil pairing. Here $3\nmid(5-1)$, so we deduce $k=1$. $\endgroup$ – Jyrki Lahtonen Dec 13 '13 at 18:40
  • 1
    $\begingroup$ @JyrkiLahtonen Yes, you are right. But here it is even easier. We see that the group has $9$ elements, and $P=(0,1)$ has order $9$. This is easy to check. So we are done. $\endgroup$ – Dietrich Burde Dec 13 '13 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy