# Clarifying a comment of Serre

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?

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

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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 $l$.