Open subgroup of $ \mathcal{O}_L $

Question

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.

Answer 1

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 $L$.