Let $\rho_{\ell}$ be the "mod $\ell$" Galois representation associated to an elliptic curve $E/K$ (i.e., corresponding to the action of Galois on the $\ell$-torsion points). Serre proved that in the case where the image of Galois is the normalizer of a nonsplit Cartan subgroup, this defines a quadratic extension of $K$ which is actually unramified.

In the course of the proof, he makes the following remark, which I cannot decipher. If $E$ has multiplicative reduction at a prime $v$ not dividing $\ell$, then the theory of Tate curves gives the exact sequence

$$ 0 \rightarrow \mu_{\ell} \rightarrow E_{\ell} \rightarrow \mathbb{Z}/\ell \mathbb{Z} \rightarrow 0, $$

which is compatible with the action of the inertia group at $v$, denoted $I_v$. Therefore, the image of $I_v$ under this Galois representations is either trivial or cyclic of order $\ell$.

Now, I see why this must be the case: since $v \nmid \ell$, the inertia acts trivially on $\mu_{\ell}$ (the $\ell$th roots of unity). According to the theory of tate curves, the $\ell$-torsion points are generated by $\mu_{\ell}$ and $q^{1/\ell}$; this is either a degree $\ell$ extension or trivial.

However, this doesn't seem to have anything to do with Serre's exact sequence, and I figure that learning how Serre sees this little fact could be useful. Can someone tell me what he means?


2 Answers 2


First a couple of references for those who may not know the source. The statement you are asking about is in page 296 of Serre's "Propriétés galoisiennes des points d'ordre fini des courbes elliptiques". The origin of the exact sequence you mentioned is explained in Serre's "Abelian l-adic representations and elliptic curves", Ch. IV, p. 31.

Now, about your question: one page prior to the statement in question (in p. 295) Serre explains that $\varphi_l(I_w)$ is of order $l-1$ or $l(l-1)$, and it can be represented as a subgroup of matrices of the form $$\left\{ \left(\begin{array}{cc}a & b \\ 0 & 1\\\end{array}\right): a\in \mathbb{F}_l^\ast, \ b\in \mathbb{F}_l\right\}.$$ The short exact sequence $0\to \mu_l \to E_l \to \mathbb{Z}/l\mathbb{Z}\to 0$, compatible with the action of inertia, says (to me anyways) that we can identify the kernel of reduction $E_l \to E \bmod l$ with $\mu_l$. The action of inertia on the kernel of reduction is given precisely by the upper left corner of the matrices given above. Since the action of inertia $I_w$ on $\mu_l$ is trivial (for the reason you mentioned, $v\nmid l$), the upper left corner of the matrix representation of the action of inertia is trivial, and we conclude that $\varphi_l(I_w)$ is a subgroup of

$$\left\{ \left(\begin{array}{cc}1 & b \\ 0 & 1\\\end{array}\right): b\in \mathbb{F}_l\right\}.$$

Thus, the image of $I_w$ is trivial or of order $l$.


As you write, when $E$ is a Tate curve, the $\ell$-torsion $E[\ell]$ contains $\mu_{\ell}$ as a Galois-sub-module, with the quotient $E[\ell]/\mu_{\ell}$ being generated by the image of $q^{1/\ell}$. Note that any Galois conjugate of $q^{1/\ell}$ is equal to $\zeta q^{1/\ell}$ for some $\zeta \in \mu_{\ell}$, and so the Galois action on $q^{1/\ell}$ modulo $\mu_{\ell}$ is trivial. This gives the exact sequence $0 \to \mu_{\ell} \to E[\ell] \to \mathbb Z/\ell \to 0.$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .