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Let $K$ be a number field and let $p$ be an odd prime. There is a natural map $$ \mathcal{O}_K \otimes_\mathbb{Z}\mathbb{Z}_p \to K \otimes_\mathbb{Q}\mathbb{Q}_p,\quad \alpha \otimes n \mapsto \alpha \otimes n, $$ and in fact the image of this map is contained in the integral closure, $A$, of $\mathbb{Z}_p$ in $K \otimes_\mathbb{Q} \mathbb{Q}_p$. I would like the map to give an isomorphism $\mathcal{O}_K \otimes_\mathbb{Z}\mathbb{Z}_p \cong A$. I originally tried using Lemma 10.147.4 of the Stacks Project, but that would require $\mathbb{Z} \to \mathbb{Z}_p$ to be a smooth ring map, which I don't think it is (although I don't really understand what a smooth ring map is). It feels like this should be true, but I have made a few attempts to prove it and not really got anywhere.

So, to be explicit, my question is: is the natural map $\mathcal{O}_K \otimes_\mathbb{Z}\mathbb{Z}_p \to K \otimes_\mathbb{Q}\mathbb{Q}_p$ an isomorphism onto the integers of $K\otimes_\mathbb{Q} \mathbb{Q}_p$?

Sebastian Monnet
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    Yes. Any element $a\in K \otimes_\mathbb{Q}\mathbb{Q}p - \mathcal{O}_K \otimes\mathbb{Z}\mathbb{Z}p$ is such that $\Bbb{Z}_p[a]$ is not a finitely generated $\Bbb{Z}_p$-module, so $a$ can't be killed by a monic $\Bbb{Z}_p[x]$ polynomial. Clearly to say that I am thinking to the valuations on $\prod_j K{\mathfrak{p}j} \cong K \otimes\mathbb{Q}\mathbb{Q}p$. It may not be immediate to you, nor that $\prod_j O{K_{\mathfrak{p}j}} $ corresponds to $O_K \otimes\mathbb{Z}\mathbb{Z}_p$ in this isomorphism. – reuns Feb 27 '22 at 13:35
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    Alternatively, if you know the map corresponds to the canonical map $\prod_jO_{K_{\mathfrak{p}j}}\rightarrow\prod_jK{\mathfrak{p}j}$ in the above notation, this reduces to noting that the integral closure of $\mathbb{Z}_p$ in $K{\mathfrak{p}j}$ is $O{K_{\mathfrak{p}j}}$, which follows since the latter is integrally closed, hence contains the former, which is a DVR by general theory, and both have field of fractions $K{\mathfrak{p}_j}$ since this is finite over $\mathbb{Q}_p$. – Thorgott Feb 27 '22 at 15:04
  • @Thorgott I do not know that fact, but I do know that the integral closure of $\mathbb{Z}p$ in $K{\mathfrak{p}j}$ is $\mathcal{O}{\mathfrak{p}_j}$, so the fact suffices to solve my problem. Could you perhaps point me to a proof of the fact or explain why it is true? – Sebastian Monnet Feb 27 '22 at 15:07
  • Maybe this question helps. – Thorgott Feb 27 '22 at 15:17

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