Let $K$ be a number field and $L/K$ an algebraic Galois extension. If $v_{P}$ is the place of $K$ corresponding to the prime ideal $P$, then we know that there are valuations of $L$ extending $v_{P}$. For example, in the finite case, we can look at the factorization $$PO_L=Q_1^{e_1}\cdots Q_r^{e_r}$$ to get the valuations $v_{Q_i}$ extending $v_{P}$. We also know that al these valuations $v_{Q_i}$ are conjugate. My question is: are all the completions $L_{v_{Q_i}}$ isomorphic? I believe that the answer is positive ("philosophically"), since when one consider $\mathbb{Q}$ and $\bar{\mathbb{Q}}$, take a prime $p$, and get $\mathbb{Q}_p$ and $\bar{\mathbb{Q}}_p$, this last one as completion of $\bar{\mathbb{Q}}$ with respect to a place over $p$, this last place over $p$ is not explicitly specified, and $\bar{\mathbb{Q}}_p$ coincides with $\overline{\mathbb{Q}_p}$, the algebraic closure of $\mathbb{Q}_p$.
I have not the right idea to prove this fact. Can you give me an hint, or maybe a reference? What can we say in the non-Galois case, when the valuation extending are not conjugate anymore?
Sure the completion is $$O_{K,Q_1}=\varprojlim_{n\to \infty} O_K/Q_1^n\cong \varprojlim_{n\to \infty} \sigma_j(O_K)/\sigma_j(Q_1^n)=\varprojlim_{n\to \infty} O_K/Q_j^n$$
$\sigma_j$ doesn't induce an automorphism of $O_{K,Q_1}$, it is an isomorphism between $O_{K,Q_1}$ and $O_{K,Q_j}$ revealing two different embeddings of $O_K$ into $\overline{\Bbb{Q}_p}$.
Try with $p=5,O_K = \Bbb{Z}[i], Q_1=(2+i),Q_2=(2-i)$. Then $\sqrt{-1}=\frac12 \sqrt{1-5}=\frac12 \sum_{k\ge 0} {1/2\choose k} (-5)^k$ is in $\Bbb{Z}_5$ and depending on if we identify $i$ with $\sqrt{-1}$ or $-\sqrt{-1}$ we get two different $5$-adic valuations on $\Bbb{Z}[i]$.