# Galois representations and isogenies of elliptic curves

Let $E$ be an elliptic curve over $\mathbb{Q}$. For each prime $\ell$, the action of $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ on $E[\ell]$ (the group of $\ell$-division points of $E$) defines a representation $$\rho=\rho_\ell:\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}) \longrightarrow \mathrm{GL}(2,\mathbb{F}_\ell).$$

Then how to prove that $\rho$ is reducible if and only if $E$ admits an isogeny of degree $\ell$?

The Galois representation is reducible iff there is a one-dimensional Galois-stable $\mathbb{F}_{\ell}$-subspace, say $C$. Then $E \rightarrow E/C$ is a $\mathbb{Q}$-rational isogeny. Conversely, if $E \rightarrow E'$ is a degree $\ell$ isogeny, its kernel $C$ is a Galois-stable subgroup of $E(\overline{\mathbb{Q}})$ of order $\ell$ so gives a one-dimensional Galois-stable $\mathbb{F}_{\ell}$-subspace of $E[\ell]$.
• @Andrew: Yes. See for instance Proposition III.4.12 of Silverman's Arithmetic of Elliptic Curves. Or, for curves over $\mathbb{Q}$, the quotient is certainly a curve and one can compute its genus by looking over $\mathbb{C}$ and this can be done either by the Riemann-Hurwitz formula or by thinking in terms of uniformizing lattices. Commented Mar 17, 2015 at 1:35
• Another reason why $E/C$ is elliptic: you have $E\rightarrow E/C\rightarrow E$, the composition being multiplication by $p$. The genus can not increase at any arrow, but this forces the genus of $E/C$ not to be zero. Commented Mar 17, 2015 at 16:34
• @Lubin: Right, and "the genus cannot increase at any arrow" by Riemann-Hurwitz. (I did have this argument in mind.) There are other ways to use Riemann-Hurwitz as well: e.g. viewing $C$ as a group of deck transformations of $E$, the quotient map is a (geometrically!) Galois covering. Commented Mar 17, 2015 at 17:50