# Higher ramification groups of $\mathbb{Q}_p(\zeta_{p^a})$

I'm trying to figure out how to compute higher ramification groups of $\mathbb{Q}_p(\zeta_{p^a})$. I've tried to proceed as follows. Write $L= \mathbb{Q}_p(\zeta_{p^a})$ and $K=\mathbb{Q}_p$, so we know that $G=\textrm{Gal}(L/K)\simeq (\mathbb{Z}/p^a\mathbb{Z})^\times$. Since the extension is totally ramified we immediately have

$$G_{-1}=G,\;\; G_0=G$$

and we also know that $G_1$ is the $p$-Sylow subgroup of $G$. To compute $G_i$ for $i\geq 1$ we know that we have an injective homomorphism given by

$$G_i/G_{i+1}\hookrightarrow U_L^{(i)}/U_L^{(i+1)}\simeq \overline{L},\;\;\sigma \mapsto \frac{\sigma \pi_L}{\pi_L}$$

where the uniformizer of $L$ is $\pi_L=1-\zeta_{p^a}$. Writing $\sigma_t$ for the element of $G$ s.t. $\zeta_{p^a}\mapsto \zeta_{p^a}^t$ we have that $\sigma_t\in G_i$ satisfies

$$\sigma_t\in G_{i+1}\Leftrightarrow \frac{\sigma_t \pi_L}{\pi_L}=\frac{1-\zeta_{p^a}^t}{1-\zeta_{p^a}}=1+\zeta_{p^a}+\ldots+\zeta_{p^a}^{t-1}\equiv 1\,(\textrm{mod }\pi_L^{i+1})$$

I'm stuck here for the case when $i>0$, since $\pi_L^{i+1}$ seems like an annoying element to work with and I can't seem to figure out how to represent $\zeta_{p^a}$ modulo $\pi_L^{i+1}$.

Any ideas on how to proceed? I'm not necessarily looking for a complete solution more like hints. I'm also curious if there are any other approaches to this problem that might work as I'm interested in general tricks that apply to computing higher ramification groups.

EDIT: I looked at Neukirch and he assigns the same problem at the end of chapter 2. Up to that point the book has covered nothing except classical theory of number fields and basic theory of valued fields. I'm more or less looking for hints in this context, though I can come back to more advanced hints once I have time to study that theory.

• Here's a (possibly rather obscure) hint: the Artin conductor of a character of G is going to be exactly what you think it is! – David Loeffler Jan 2 '12 at 11:44
• @DavidLoeffler: Yes, the text that I'm studying doesn't even define the Artin conductor, so that's definitely a bit obscure. :) – pki Jan 2 '12 at 11:53
• See en.wikipedia.org/wiki/Artin_conductor. From my (rather biased) viewpoint I'd say that defining the Artin conductor is one of the most important applications of higher ramification theory, so it's bizarre that your text doesn't mention it. – David Loeffler Jan 2 '12 at 11:54
• It's not a textbook it's a lecture note on the theory of local fields. – pki Jan 2 '12 at 11:55
• I tried to keep my hint relatively minimal. Let me know if you'd like more details. – Cam McLeman Jan 3 '12 at 16:35

A hint: Let $v_L$ denote the valuation in $L$. Then just by clearing denominators in your formula, you have $$\sigma_t\in G_i \iff v_L(\sigma_t(\pi_L)-\pi_L)\geq i+1.$$ Now given your explicit form of $\pi_L$ and $\sigma_t$ (long live cyclotomic extensions!), it's not hard to determine this valuation -- it's now reduced to determining $i$ in terms of orders of $t$ modulo various powers of $p$ less than $p^a$, etc.
• I was hoping you could add some more details, in particular how to handle divisibility by $\pi_{L}^{i}$ for $i \geq 2$. – Jacob Bond Feb 29 '16 at 4:06
The complete answer can be found as a developped example in Serre's "Local Fields", proposition 18 at the end of chapter 4. Since you seem to be interested in "hints" rather than in detailed solutions, here is Serre's hint (added after the proof of his proposition), which comes from Class Field theory. Consider the extension $L_\infty$ obtained by adding to $\mathbf Q_p$ all the $p^a$-th roots of 1. By the (elementary) theory of cyclotomic fields, the Galois group $G_\infty$ of $L_\infty /\mathbf Q_p$ is canonically isomorphic to the multiplicative group $\mathbf Z_p^*$ of $p$-adic units. The latter group is naturally filtered by the subgroups $U_r=1+p^r\mathbf Z_p$ ; extend this filtration to non integral indices $v$ by defining $U_v = U_r$ if $r$ is the smallest integer $\ge v$. Then, by CFT (see Serre's chapter 15), the canonical isomorphism $\mathbf Z_p^* \cong G_\infty$ transforms the filtration $U_v$ of $\mathbf Z_p^*$ into the filtration of $G_\infty$ consisting of the ramification subgroups in the upper indexation. Knowing what the result must look like, it is easy to prove propos. 18 by a direct computation starting from the definition of the ramification indices (in the lower indexation) - you may consider this as the missing details in @Cam McLeman .