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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 $v \nmid \ell$?

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user261123
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    Do you know where the number field $\mathbb Q(\zeta_\ell)$ is ramified? – Bruno Joyal Sep 08 '15 at 22:12
  • Yes only in $\ell$, since the inertia subgroup has cardinality equal to the ramification index then can be only the identity over the other prime. Is this right? – user261123 Sep 09 '15 at 06:10
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    Well the argument isn't right, but you are right about it being ramified only in $\ell$. From this can you deduce something about the action of Galois on $\mu_\ell$? – Bruno Joyal Sep 09 '15 at 15:14
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    BTW your question will have more chance of being answered if you ask only one question per post. Otherwise it's more work for people to answer. – Bruno Joyal Sep 09 '15 at 15:14
  • You are right about the number of question but I would not annoying people with too much post. For your second hint, I know that the Galois acts permuting the element of $\mu_{\ell}$ but don't think this is the point. – user261123 Sep 09 '15 at 15:28

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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 $p\neq \ell$.

Álvaro Lozano-Robledo
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