# Open subgroup of units of local field (norm index computation)

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?

• $\exp(x) = 1+x+O(x^2)$ 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. Jun 11, 2019 at 21:39
• @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 $\exp(\pi^n\mathcal{O}_K) = 1 + \pi^n\mathcal{O}_K$? Jun 11, 2019 at 21:47
• 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) Jun 11, 2019 at 21:48
• 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. Jun 12, 2019 at 15:42
• @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). Jun 14, 2019 at 9:42