I know that this follows from the existence and non-degeneracy of the Weil pairing. A consequence of the existence of the Weil pairing is that, if the whole $n$-torsion is defined over $\mathbf Q$, then the $n$-th roots of unity $\mu_n \subseteq \mathbf Q$. This of course implies $n=2$. Thus, for an odd prime $p$, $E(\mathbf Q)[p]$ is cyclic, being a proper subgroup of $\mathbf Z/p \times \mathbf Z/p$. Putting this information together for the various primes dividing $n$ shows that for a general odd $n$, $E(\mathbf Q)[n]$ is cyclic.

But this seems to me like over-kill. Could this simple fact be proven without using the Weil pairing?

  • $\begingroup$ IDK, but I rather think that this argument using the Weil pairing is quite elegant. It also shows that if we look at the torsion points of an EC over a finite field $\mathbb{F}_q$, then the torsion group is cyclic whenever $\gcd(n,q-1)=1$. $\endgroup$ – Jyrki Lahtonen Feb 23 '13 at 19:57

Here's an analytic/topologic proof (also you can replace $\Bbb Q$ with any field contained in $\Bbb R$) :

Look at $E(\Bbb R)$ sitting inside the complex torus $E(\Bbb C) \cong \Bbb C / \Lambda$. Topologically, it has dimension $1$ (it locally looks like a line), and it has one or two connected components (depending on the number of real roots of $x^3- g_2x - g_3$ when you put the curve in Weierstrass normal form)

Pick a fundamental parallelogram for $\Bbb C/\Lambda$ such that the connected component of $0$ is a side of the domain (this amounts to choosing a nice basis for $\Lambda$). Then it is clear that $E(\Bbb R) \cong \Bbb R/\Bbb Z$ if there is one component, and $E(\Bbb R) \cong \Bbb R/\Bbb Z \times \Bbb Z /2\Bbb Z$ if it has two.

Now, you simply need to check that the $n$-torsion of those two groups is cyclic, and since $E(\Bbb Q)$ is a subgroup of $E(\Bbb R)$, its $n$-torsion is a subgroup of that cyclic group, hence it is cyclic.

  • $\begingroup$ Ah, of course! Thank you mercio. (I'll leave the question open for a little longer to see if there are other ideas.) $\endgroup$ – Bruno Joyal Feb 23 '13 at 20:08

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