# Kernel of a 3-isogeny between two elliptic curves

Suppose $E_1$ and $E_2$ are two elliptic curves defined over $\mathbb{Q}$ and there exists a 3-isogeny $\varphi$: $E_1 \longrightarrow E_2$. 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 $?$

• What is a $3$-isogeny? – abx Mar 30 '14 at 18:41
• 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 $\mu _3$. – abx Mar 30 '14 at 19:31
• 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 $\mathbb{Q}(e^{2\pi i/3} )$. – abx Mar 30 '14 at 20:04
• 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. – Chris Wuthrich Mar 30 '14 at 22:23
• 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 $\hat E(\mathbb{Q})$. – Joe Silverman Mar 30 '14 at 22:55

The simplest example is $y^2=x^3+D$, which has a 3-isogeny 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.
• There are probably explicit formulas for 3-isogenies 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. – Joe Silverman Mar 31 '14 at 11:15