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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?

Lios
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  • To "know" that those valuations are conjugate, you need to assume something like $L/K$ being a Galois extension. Are you assuming this? – Angina Seng May 16 '20 at 10:07
  • You're absolutely right, I edit. – Lios May 16 '20 at 10:17
  • @Lios : Refer to my question : https://math.stackexchange.com/questions/4954735/for-galois-extension-of-number-fields-mathfrakp-j-mathfrakp-k-mathfr – Plantation Aug 05 '24 at 11:55

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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 $\Bbb{Z}[i]$.

reuns
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    Ok let me check if it is clear to me. You are saying that to show if the completion of the field are isomorphic it is enough to show that their valuations ring are isomorphic, since the local fields are just the quotient fields of their valuation rings. – Lios May 16 '20 at 17:53
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    So since $\sigma_j$ is an isomorphism and send $Q_1$ to $Q_j$, we get the isomorphism between the completions of the ring of integers induced by $\sigma_j$. Therefore also the completions of the fields are isomorphic. – Lios May 16 '20 at 17:53
  • Uhm..What is exact definition of $\mathcal{O}{K, Q_1} $? Completion of $\mathcal{O}_K$ at $Q_1$ ? It seems that $Q_1$ is prime ideal of $\mathcal{O}_L$. The completion $\mathcal{O}{K,Q_1}$ makes sense? Which valuation is the completion about? And why $\mathcal{O}{K , Q_1} \cong \mathcal{O}{K,Q_j}$ implies $L_{Q_1} \cong L_{Q_j}$? $\mathcal{O}{K,Q_1}$ ( resp. $\mathcal{O}{K, Q_j}$ ) are the valuation rings of $L_{Q_1}$ ( resp. $L_{Q_j}$ ) ? – Plantation Aug 05 '24 at 09:45