Let $ L/ K $ be a finite Galois extension with $ K $ a local field and $ G $ its Galois group. By the normal basis theorem, there is a normal basis $ \{ \sigma_i(x) : \sigma_i \in G \} $, moreover, by multiplying by an appropriate element $ d \in \mathcal{O}_K $, we can assume that these elements are in $ \mathcal{O}_L $. My question is: why is the subgroup $ \sum_i \mathcal{O}_K \sigma_i(x) $ open in $ \mathcal{O}_L $? This is probably very obvious, but I'm not sure what's the answer.
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If $v_1, ... , v_n$ is a basis for $L$ over $K$, then $(c_1, ... , c_n) \mapsto c_1v_1 + \cdots + c_nv_n$ defines a homeomorphism $K^n \rightarrow L$. So the image of $\mathcal O_K^n$ is open in $L$.
D_S
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Why is $\mathcal O _K$ open in $L $? – mathemather Jul 19 '20 at 12:28
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It is never open in $L$ if $K \subsetneq L$ – D_S Jul 19 '20 at 21:34