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let $K / \mathbb{Q}$ be a Galois number field with ring of integers $O_K $. take a prime ideal $\mathfrak P$ of $O_K$ lying over $p\mathbb{Z}$. then we obtain the induced field extension $K_{\mathfrak P} /\mathbb{Q}_p$ with $K_{\mathfrak P}$ completion of $K$ with respect $\mathfrak P$. why is this extension also Galois?

since $\mathbb{Q}_p$ has characteristic zero the induced extension is separable. what about normal? since $K / \mathbb{Q}$ normal because Galois we can find a family $\{f_i \}_{i \in I}$ of polynomials with coefficients in $\mathbb{Q}$ such that $K$ is nothing but their splitting field. by construction all $\{f_i \}_{i \in I}$ split also in $K_{\mathfrak P}$ since $K \subset K_{\mathfrak P}$. unfortunately I don't see how to continue from here...

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    Write $K = \mathbf{Q}[x]/p(x)$, so $p(x)$ is separable. Over $\mathbf{Q}p$, the polynomial $p(x)$ factors as $\prod p_i(x)$, and the fields $K{\mathbf{P}i}$ are none other than the fields $\mathbf{Q}_p[x]/p_i(x)$. (Equivalently, $K \otimes \mathbf{Q}_p = \prod K{\mathfrak{P}i}$). To show that $K_i = K{\mathfrak{P}_i}$ is normal, it suffices to show that $K_i$ contains all the roots of $p_i(x)$ and is thus its splitting field. But this field contains $K$, and thus (since $K$ is a splitting field) it also contains all the roots of $p(x)$, which includes the roots of $p_i(x)$ for every $i$. – The Piper Jun 29 '19 at 15:40
  • @The Piper: why the completion $K_{P_i}$ equals $\mathbb{Q}_p[x]/p_i(x)$? –  Jun 29 '19 at 15:55
  • If $p = \prod \mathfrak{p}^{e_i}i$ in a number field $K$, then $\mathcal{O}_K/p^n \simeq \prod \mathcal{O}_K/\mathfrak{p}^{n e_i}_i$. Can you see from this that $\mathcal{O}_K \otimes \mathbf{Z}_p = \prod \mathcal{O}{K,\mathfrak{p}i}$ and thus $K \otimes \mathbf{Q}_p = \prod K{\mathfrak{p}_i}$? – The Piper Jun 29 '19 at 16:03
  • @ThePiper: the last two aren't clear. $\mathcal{O}K \otimes \mathbf{Z}_p$ is just the localization wrt the multiplicative system $O_K \backslash p$. how do you obtain the direct product on the right side? the conclusion $K \otimes \mathbf{Q}_p = \prod K{\mathfrak{p}_i} $ is also unclear –  Jun 29 '19 at 16:24
  • If they are unclear, then I suggest spending time thinking about them some more, because they should be transparent. – The Piper Jun 29 '19 at 16:30
  • @ThePiper: ok, regarding $\mathcal{O}K \otimes \mathbf{Z}_p = \prod \mathcal{O}{K,\mathfrak{p}i}$: obviously we have an inclusion between the multiplicative systems $S_i:= O_K \backslash \mathfrak{p_i} \subset O_K \backslash p:= S$ this induces $O{K,\mathfrak{p_i}} \subset O_{K,p}= \mathcal{O}K \otimes \mathbf{Z}_p$. Do you mean that $\mathcal{O}_K \otimes \mathbf{Z}_p = \prod \mathcal{O}{K,\mathfrak{p}_i}$ as $\mathbf{Z}_p$-modules or rings? –  Jun 29 '19 at 17:08
  • since in the category of modules finite products and sums coinside so $\prod \mathcal{O}{K,\mathfrak{p}_i}=\oplus \mathcal{O}{K,\mathfrak{p}i}$ as modules and by universal property of direct sum/coproduct we obtain $\oplus \mathcal{O}{K,\mathfrak{p}_i} \to O_K \otimes \mathbf{Z}_p$. since all $\mathfrak{p_i}$ are maximal it should be (I have to think a bit more why) an isomorphism of $\mathbb{Z}_p$-modules. –  Jun 29 '19 at 17:08
  • question: do you mean the identification $\mathcal{O}K \otimes \mathbf{Z}_p = \prod \mathcal{O}{K,\mathfrak{p}_i}$ as modules or as rings? regarding "ring" case I have now idea what to do. –  Jun 29 '19 at 17:09
  • As rings --- start with $\mathcal{O}_K/p^n = \prod \mathcal{O}_K/\mathfrak{p}^{n e_i}_i$ as mentioned then take inverse limits in $n$. – The Piper Jun 29 '19 at 18:44
  • I see... then on the left we observe that since inverse limits behave well under tensor extensions (in the sense $\varprojlim B/p^n \cong B \otimes \varprojlim A/pA$)and on the right hand side the ${ \prod \mathcal{O}K/\mathfrak{p}^{n e_i}_i }_n$ form a cofinal system of $\prod \mathcal{O}{K,\mathfrak{p}_i}$ so limits coinside. –  Jun 29 '19 at 19:14
  • could you give a hint for $K \otimes \mathbf{Q}p = \prod K{\mathfrak{p}_i}$? –  Jun 29 '19 at 19:14
  • $(p) = (\prod_j P_j)^e$ then $ \varprojlim O_K/(p)^n = \varprojlim O_K/(\prod_j P_j)^{en} = \varprojlim \prod_j O_K/P_j^{en}=\prod_j \varprojlim O_K/P_j^n = \prod_j O_{\displaystyle K_{P_j}}$ – reuns Jun 29 '19 at 19:44
  • this gives the equation $\mathcal{O}K \otimes \mathbf{Z}_p = \prod \mathcal{O}{K,\mathfrak{p}i} $ for the rings of intergers. how does it help to obtain $K \otimes \mathbf{Q}_p = \prod K{\mathfrak{p}_i} $? –  Jun 29 '19 at 21:23
  • Tim, from $K=\Bbb{Q}[x]/\langle p(x)\rangle$ you immediately get $K\otimes\Bbb{Q}p=\Bbb{Q}_p[x]/\langle p(x)\rangle$ (vector spaces are flat so tensoring with $\Bbb{Q}_p$ preserves quotients). Then the Chinese remainder theorem isolates the component fields $\prod K{\mathfrak{p}_i}$. See for example here – Jyrki Lahtonen Jun 30 '19 at 05:43
  • I understand that by Chinese remainder thm we obtain the splitting $K \otimes \mathbb{Q}p \cong \prod\limits{i=1}^g K_{p_i}$. CRT gives $K_{p_i}=\mathbf{Q}p[x]/p_i(x)$ and the discussion above donates us $\mathcal{O}_K \otimes \mathbf{Z}_p = \prod \mathcal{O}{K,\mathfrak{p}i}$ with $K{\mathfrak{P}i}= Frac(\mathcal{O}{K,\mathfrak{p}i})$ by proceeding to limit & using cofinal argument. the problem is still to derive from this that then the $K{p_i}$ already coincide with completion $K_{\mathfrak{P}_i}$ of $K$ wrt $p_i$. –  Jul 01 '19 at 09:59
  • unfortunately I still uncertain how we conclude $K \otimes \mathbf{Q}p = \prod K{\mathfrak{P}i}$. again to clarify the notations the polynomial $p(x)$ splits to $\prod p_i(x)$ over $\mathbb{Q}_p[X]$. we denote by $K{p_i}=\mathbf{Q}p[x]/p_i(x)$ and by $K{\mathfrak{P}i}$ the completion of $K$ with respect the prime ideal $\mathfrak{P}_i:=(p_i)$. we know by CRT that $K \otimes \mathbb{Q}_p \cong \prod\limits{i=1}^g K_{p_i}$ and by taking inverse limits we also saw –  Jul 05 '19 at 01:54
  • that $\mathcal{O}K \otimes \mathbf{Z}_p = \prod \mathcal{O}{K,\mathfrak{P}i}$ where $\mathcal{O}{K,\mathfrak{P}i}$ are the rings of integers of $K{\mathfrak{P}i}$. why is that sufficient to obtain $K{\mathfrak{P}i}= K{p_i}$ or equivalently $K \otimes \mathbf{Q}p = \prod K{\mathfrak{P}_i} $? sorry for frequently annoying but I just don't see final step –  Jul 05 '19 at 01:55

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