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during the discussions in this former thread there occured a statement that causes some headachse to me. we work in the same setting as given in the linked thread: 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 $\mathfrak P$.

now $pO_K$ ramifies in $O_K$ to $pO_K = \prod_i \mathfrak{P}_i ^{l_i}$. my question how to deduce that following formula is correct:

$$K \otimes \mathbf{Q}_p = \prod K_{\mathfrak{P}_i}$$

in this equation we denote by $K_{\mathfrak{P}_i}$ the completion fields of $K$ with respect to prime $\mathfrak{P}_i$.

in the progression of the discussion there were obtained some results bringing the solution nevertheless the final step combining the observations presented below is still missed:

we observed that

($\bullet$) since $K$ separable and finite because number field we can apply the primitive generator theorem and obtain $K= \mathbb{Q}[X]/R(X)$ for $R(X) \in \mathbb{Q}[X]$. in $\mathbb{Q}_p[X]$ we obtain splitting of $R$ to $R(X) = \prod_i R_i(X)$ with irreducible and coprime $R_i(X) \in \mathbb{Q}_p[X]$. Chinese remainder theorem tell that $K \otimes \mathbf{Q}_p = \prod_i \mathbb{Q}_p[X]/R_i(X)= \prod_i K_{i}$ with $K_i := \mathbb{Q}_p[X]/R_i(X)$.

($\bullet$) on the other hand having ramification splitting $pO_K = \prod_i \mathfrak{P}_i ^{l_i}$ in $O_K$ we use again CRT in ideal theoretic way to obtain $\mathcal{O}_K/p^n \cong \prod_i \mathcal{O}_K/\mathfrak{P}^{n l_i}_i$. passing to inverse colimit and observe that the inverse system $(\prod_i \mathcal{O}_K/\mathfrak{P}^{n l_i}_i)_n$ is cofinal in bigger inverse system $(\prod_i \mathcal{O}_K/\mathfrak{P}^{n_{i} l_i}_i)_{(n_1,...,n_s)}$ we obtain $O_K \otimes_{\mathbb{Z}} \mathbb{Z}_p\cong \varprojlim_n O_K/p^n= \varprojlim_n \prod_i \mathcal{O}_K/\mathfrak{P}^{n l_i}_i= \prod _i \varprojlim_n \mathcal{O}_K/\mathfrak{P}^{n }= \prod _i O_{K_{\mathfrak{P}_i}}$ again by cofinal argument.

now I don't know haow to derive from facts that $K \otimes \mathbf{Q}_p = \prod_i \mathbb{Q}_p[X]/R_i(X)= \prod_i K_{i}$ and $O_K \otimes_{\mathbb{Z}} \mathbb{Z}_p \cong \prod _i O_{K_{\mathfrak{P}_i}}$ the equation $K \otimes \mathbf{Q}_p = \prod K_{\mathfrak{P}_i}$. is there some useful relation connection behavior of fraction fields of their rings of integers? other way to tensor $O_K \otimes_{\mathbb{Z}} \mathbb{Z}_p$ with $K$ and $\mathbb{Q}_p$ breaks in generally the product structure on the right hand side.

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    https://math.stackexchange.com/questions/137888/why-is-o-k-otimes-mathbbz-p-cong-oplus-mathfrakppo-k-mathfrakp?rq=1 related – Maxime Ramzi Jul 05 '19 at 14:20

1 Answers1

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This is how I see those things, did I miss something important ?

$pO_K = \prod_j P_j^e$ for distinct maximal ideals $P_j$. The $P_j$ are comaximal thus so are the $P_j^e$ $$O_K/(p^n) = O_K/(\prod_j P_j^{en})\cong \prod_j O_K/P_j^{en}$$

$$O_K \otimes_\Bbb{Z} \Bbb{Z}_p=O_K \otimes_\Bbb{Z} \varprojlim\Bbb{Z}/(p^n) \cong \varprojlim O_K \otimes_\Bbb{Z} \Bbb{Z}/(p^n) \cong \varprojlim O_K / (p^n)\\ \cong \varprojlim \prod_j O_K/P_j^{en} \cong \prod_j \varprojlim O_K/P_j^{en} = \prod_j O_{K_{P_j}}$$ For that $\varprojlim $ commutes with $\otimes_\Bbb{Z}$ I would use a free $\Bbb{Z}$-module basis $O_K= \sum_{r=1}^N \beta_r \Bbb{Z}$. For $\varprojlim \prod_j O_K/P_j^{en} \cong \prod_j \varprojlim O_K/P_j^{en}$ I would inject the latter in the former.

$$K \otimes_\Bbb{Q} \Bbb{Q}_p \cong O_K \otimes_\Bbb{Z} (\Bbb{Z}_p[p^{-1}])\cong (O_K \otimes_\Bbb{Z} \Bbb{Z}_p)[p^{-1}] \cong (\prod_j O_{K,P_j})[p^{-1}]\cong \prod_j ( O_{K,P_j}[p^{-1}]) = \prod_j K_{P_j}$$

(I would use $O_K = \sum_{r=1}^N \beta_r \Bbb{Z},K= \sum_{r=1}^N \beta_r \Bbb{Q}$)

Finally $K= \Bbb{Q}(\alpha) \cong \Bbb{Q}[x]/(f(x))$ and $$K \otimes_\Bbb{Q} \Bbb{Q}_p \cong \Bbb{Q}[x]/(f(x))\otimes_\Bbb{Q}\Bbb{Q}_p \cong \Bbb{Q}_p[x]/(f(x)) =\Bbb{Q}_p[x]/(\prod_i f_i(x)^d) \\ \cong \prod_i \Bbb{Q}_p[x]/(f_i(x)) \cong \prod_i K_{i}$$

Note $d=1$ because otherwise $f$ would have a double root in $\overline{\Bbb{Q}_p}$ thus it would have a double root in $\overline{\Bbb{Q}}$.

The $K_i$ and the $K_{P_j}$ are the same because ...

reuns
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  • two steps still not convince me. why $O_K \otimes_\Bbb{Z} (\Bbb{Z}p[p^{-1}])\cong (O_K \otimes\Bbb{Z} \Bbb{Z}p)[p^{-1}]$ and $O{K,P_j}[p^{-1}]=K_{P_j}$? take into account that $\Bbb{Z}_p[p^{-1}]$ isn't free $\Bbb{Z}_p$-module. could you explain these two equalities? –  Jul 08 '19 at 11:03
  • $O_{K,P_j}[p^{-1}]$ is the integral domain generated by $O_{K,P_j}$ and $p^{-1}$, if $a \in K_{P_j}$ then $a p^m \in O_{K_{P_j}}$ iff $v(p^m) \ge -v(a)$. That $K \otimes_\Bbb{Q} Q_p \cong (O_K \otimes_\Bbb{Z} \Bbb{Z}p)[p^{-1}]$ seems obvious to me : inject the RHS in the LHS, use that the same basis works for both $O_K = \sum{r=1}^N \beta_r \Bbb{Z},K= \sum_{r=1}^N \beta_r \Bbb{Q}$ – reuns Jul 08 '19 at 11:12
  • yes I see. thank you a lot –  Jul 08 '19 at 11:22
  • ...I think that $K_i$ and $K_{P_j}$ are the same since $K_i \subset K \subset K_{P_j}$ since $f_i$ splits in $K$ as $K$ Galois and by comparing the factors of $\prod_j K_{P_j}= \prod_i K_{i}$ we are done –  Jul 08 '19 at 11:44