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I am trying to understand following lemma from Milne's Class Field Theory: https://www.jmilne.org/math/CourseNotes/CFT.pdf#X.3.2.3 (the link will take you directly to the said lemma)

Let $L$ be a finite Galois extension of $K$ with Galois group $G$. Then there exists an open subgroup $V$ of $\mathcal O _L$ stable under $G$ such that $H^r(G,V) =0$ for all $r > 0$.

In the proof given, we obtain, using the Normal basis theorem, basis $\{x_{\tau}\in \mathcal O _L |\tau \in G\}$ for $L/K$. We define $$V=\sum_{\tau \in G } \mathcal O _K x_{\tau} $$ I want to see that this subgroup $V$ is open in $\mathcal O _L $. Here is my attempt:
Each $ x_\tau=u_\tau \pi_L^{i_\tau}$ where $u_\tau$ is a unit in $\mathcal O_L$ and $\pi _L $ its uniformizer. Define $i=\text{max}\{i_\tau\ | \ \tau \in G \} $. Then for some fixed $0<c <1 $, consider $U=B(0,c^{i}) $.
What I would like to claim is that $U \subset V $, as if $0$ has open nbhd contained in $V$ then any point has, and so $V$ is open. But what I observe is that, in fact, we can fit $V$ inside ball of any radius by multipying the $x_\tau $ by $\pi _K $. I wonder if this is somehow useful?

Feel free to give any reference. Any help is appreciated.

Update: This question was asked here: Open subgroup of $ \mathcal{O}_L $
But I don't understand why is $\mathcal O _K$ open in $L $?

mathemather
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Note that the $x_\tau$ form a $K$-basis of $L$. For each $a\in L$ there are unique $b_\tau$ in $K$ with $a=\sum_\tau b_\tau x_\tau$. Each $K$-linear map $\phi_k:a\mapsto b_\tau$ is continuous.

Now $\mathcal{O}_L$ is compact, so each $\phi_\tau(\mathcal{O}_L)$ is bounded. There are only finitely many $\tau$, so there is a in integer $M\ge0$ such that $\phi_\tau(\mathcal{O}_L) \subseteq\pi^{-M}\mathcal{O}_K$ where $\pi$ is a uniformiser in $K$. Let $U=\pi^M\mathcal{O}_L$. Then $U$ is an open subgroup of $\mathcal{O}_L$. Also each element of $U$ equals $\pi^M a=\sum_{\tau}\phi_\tau(a)x_\tau\in V$ where $a\in\mathcal{O}_L$ and $V$ is your $V$. So $U\subseteq V$. As $U$ is a subgroup of $V$, and $U$ is open, then so is $V$ (it's a union of cosets of $U$).

Angina Seng
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