Tate curve and action of inertia group I read the answers to this question Clarifying a comment of Serre. However I miss a passage of the second answer and since I can't comment there I have should post a new question. 
I don't understand the following passage
Why the action of the inertial subgroup is trivial over $\mu_{\ell}$ whenever $v \nmid \ell$?
 A: First, a quick summary of splitting of primes in Galois extensions (see here, or Marcus' Number Fields for more on this topic). Suppose that $L$ is a finite Galois extension of $\mathbb{Q}$, and let $p$ be a rational prime. Let $\mathcal{O}_L$ be the ring of integers of $L$. Then,
$$p\mathcal{O}_L = (\wp_1\cdots\wp_r)^e$$
where $\wp_i$ are the prime ideals of $\mathcal{O}_L$ lying above $p$, and $r,e\geq 1$ are integers, and $e$ is called the ramification index. The degree of the extension of finite fields $[\mathcal{O}_L/\wp_i : \mathbb{Z}/p\mathbb{Z}]$ does not depend on $i$, and we call it $f$. Now let $G=\operatorname{Gal}(L/\mathbb{Q})$ be the Galois group of the extension, fix $1\leq i\leq r$, let $\wp=\wp_i$, and let $D_\wp$ and $I_\wp$ be, respectively, the decomposition group and inertia subgroups of $G$ attached to the prime $\wp$. The size of $I_\wp$ is $e$, and the size of $D_\wp$ is $ef$. There is an inclusion of fields
$$\mathbb{Q} \subseteq L^{D_\wp} \subseteq L^{I_\wp} \subseteq L$$
where $L^H$ is the subfield of $L$ fixed by a subgroup $H$ of $G$. In particular, $[L:L^{I_\wp}]=e$.
Thus, $p$ ramifies in $L/\mathbb{Q}$ if and only if $e=1$, if and only if $I_\wp$ is trivial, if and only if $L^{I_\wp}=L$, if and only if $I_\wp$ acts trivially on all of $L$.
In your case, $L=\mathbb{Q}(\zeta_\ell)$, where $\ell$ is a prime different from $p$. Since only $\ell$ is ramified in $L/\mathbb{Q}$, it follows that $I_\wp$ acts trivially on $\mathbb{Q}(\zeta_\ell)$, for any prime ideal $\wp$ above $p\neq \ell$.
