Suppose $E_1$ and $E_2$ are two elliptic curves defined over $\mathbb{Q}$ and there exists a 3isogeny $\varphi$: $E_1 \longrightarrow E_2$. If $E_1$ has no $\mathbb{Q}$rational point of order 3, then what is Ker $\varphi$ $?$ Is it $\mu_3$ or $\mathbb{Z}/3\mathbb{Z}$ where $\mu_3$ denotes the 3rd roots of unity $?$

$\begingroup$ What is a $3$isogeny? $\endgroup$ – abx Mar 30 '14 at 18:41

1$\begingroup$ Then you have no choice! There are only two possible actions of $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ over $\mathbb{Z}/3$, either trivial (corresponding to $\mathbb{Z}/3$) or through a character of order 2, giving $\mu _3$. $\endgroup$ – abx Mar 30 '14 at 19:31

2$\begingroup$ Sorry, I wrote too hastily. In fact your group is determined by a homomorphism $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow \{\pm 1\} $, and all such homomorphisms are possible. This means that your kernel is defined by a quadratic extension of $\mathbb{Q}$, but all such extensions are allowed. If you want $\mu _3$ your kernel must be defined over $\mathbb{Q}(e^{2\pi i/3} )$. $\endgroup$ – abx Mar 30 '14 at 20:04

1$\begingroup$ It is easy to check that $\ker \varphi =\mu_3$ by checking that the dual isogeny has a rational $3$torsion point in its kernel. $\endgroup$ – Chris Wuthrich Mar 30 '14 at 22:23

1$\begingroup$ To expand on Chris's comment, the Weil pairing shows that $\operatorname{ker}(\phi)$ and $\operatorname{ker}(\hat\phi)$ are dual as Galois modules, so $\operatorname{ker}(\phi)=\mu_3$ if and only if $\operatorname{ker}(\hat\phi)=\mathbb{Z}/3\mathbb{Z}$, and the latter is equivalent to the points in the kernel of $\hat\phi$ being in $\hat E(\mathbb{Q})$. $\endgroup$ – Joe Silverman Mar 30 '14 at 22:55
The simplest example is $y^2=x^3+D$, which has a 3isogeny defined over $\mathbb{Q}$ whose kernel is $\{O,(0,\sqrt{D}),(0,\sqrt{D})\}$. As abx pointed out, as a Galois module this kernel is defined by the quadratic extension $\mathbb{Q}(\sqrt{D})$, which can be any quadratic extension. In general, if the kernel is $\{0,P,P\}$, then Galios acts via $\sigma(P)=\chi(\sigma)P$, and $\chi$ is the quadratic character in abx's comment.

$\begingroup$ How do you find out the 3isogeny defined over Q and its kernel ? $\endgroup$ – Andrew Mar 31 '14 at 11:12

$\begingroup$ There are probably explicit formulas for 3isogenies written down somewhere. Or you can use the method in Velu's paper, which works for cyclic $N$isogenies (although of course it becomes infeasible in practice if $N$ gets too large). Or you can probably work it out yourself with a bit of work; I did the (admittedly easier) two isogeny case in my Arithmetic of Elliptic Curves book. $\endgroup$ – Joe Silverman Mar 31 '14 at 11:15

1$\begingroup$ You can find explicit formulas for 3isogenies in the paper by Cohen and Pazuki which is available at the link here. $\endgroup$ – Jeremy Rouse Mar 31 '14 at 15:16