Clarifying a comment of Serre

Solution 1:

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 $l$.

Solution 2:

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 $0 \to \mu_{\ell} \to E[\ell] \to \mathbb Z/\ell \to 0.$