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Let $k$ be a $\mathfrak{p}$-adic field, $K/k$ be a cyclic extension with Galois group $G = \langle \sigma \rangle$, and $U_K$ be the group of units in (the valuation ring of) $K$. One of the results of Chapter IX of Lang's Algebraic Number Theory is the computation of the Herbrand quotient $Q(G, U_K) := (U_k : N^K_kU_K)/(\ker N^K_k : (1-\sigma)U_K)$ where $\ker N^K_k$ is the kernel of the norm as a homomorphism from $U_K$ to $U_k$. Knowing the value of this quotient is useful, e.g. in computing the index $(k^* : N^K_kK^*) = [K:k]$.

The trick Lang uses is as follows (from p. 188):

Let $\{\omega_\tau\}$ be a normal basis for $K$ over $k$. After multiplying the elements of this basis by a high power of a prime element $\pi$ in $k$, we can assume that they have small absolute value. Let $M = \sum_{\tau \in G} \mathcal{O}_k\cdot \omega_\tau$.

As far as I can tell, the reason for this first step is to ensure that the exponential map is well-defined on $M$, since the power series defining it only converges on sufficiently small neighborhoods of $0$.

The $G$ acts on $M$ semilocally, with trivial decomposition group. Furthermore, $\exp M = V$ is $G$-isomorphic to $M$ (the inverse is given by the log), and $V$ is an open subgroup of the units, whence of finite index in $U_K$. Therefore $1 = Q(G, V) = Q(G, U_K)$.

It's clear to me why this implies that $V$ has finite index ($U_K$ is compact). However, I don't understand exactly why $V$ is open in $U_K$. Should this be obvious? Also, why does this depend on $G$ being cyclic?

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  • $\begingroup$ $\exp(x) = 1+x+O(x^2)$ this guanrantees for $n$ large enough $\exp(1+\pi^n O_K) = 1+\pi^n O_K$ which is finite index open subgroup of $U_K = \langle \zeta\rangle (1+\pi O_K)$. What does it mean G acts on M semilocally, with trivial decomposition group ? Can you elaborate on $Q(G,V)$ and $1 = Q(G, V) = Q(G, U_K)$ ? Also finite index implies open. $\endgroup$
    – reuns
    Jun 11, 2019 at 21:39
  • $\begingroup$ @reuns: Oops, in the question I meant that it was clear why the index is finite from the fact that it is open, but that I wasn't sure about why it is open in the first place (the fact that it is open is the only part I don't know how to prove; the rest of the question is just providing details). Also, did you mean to write $\exp(\pi^n\mathcal{O}_K) = 1 + \pi^n\mathcal{O}_K$? $\endgroup$
    – babu_babu
    Jun 11, 2019 at 21:47
  • $\begingroup$ Yes. So can you elaborate on those $Q(G,U_K)$ and what are they useful for (the proof of the surjectivity of $\exp(\pi^n\mathcal{O}_K) \to 1 + \pi^n\mathcal{O}_K$ is the same as in Hensel lemma : fix the p-adic digits of the LHS one by ones to fit the RHS) $\endgroup$
    – reuns
    Jun 11, 2019 at 21:48
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    $\begingroup$ I think it's obvious. Prove for yourself: If $K\vert k$ is a finite extension of local fields, $\mathcal{O}_k$ the ring of integers of $k$, and $e_1, ..., e_n$ any vector space basis of $K$ over $k$, the lattice ($\mathcal{O}_k$-module) $\sum_1^n \mathcal{O}_k \cdot e_i$ is open. $\endgroup$ Jun 12, 2019 at 15:42
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    $\begingroup$ @TorstenSchoeneberg: Oops, you are right; I am very bad. Actually $K$ is a finite-dimensional normed vector space over $k$, and since $k$ is complete the norm on $K$ is equivalent to the sup norm with respect to the basis $\{e_i\}$. From that it's obvious that $M$ is open (it's just an open cube under the sup norm). $\endgroup$
    – babu_babu
    Jun 14, 2019 at 9:42

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