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?