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Suppose we have a number field $K$ with ring of integers $\mathcal O_K$. Let $\frak p$ be a prime ideal in $K$ lying over $p\in \mathbb Q$. Then, using the $\frak p$-adic norm, we may define the $p$-adic ring of integers $\mathbb Z_{\frak p} = \{ x\in K: |x|_{\frak p}\leq 1\}$. I know that $\mathcal O_K \not= \mathbb Z_{\frak p}$, but is $\mathcal O_K \subset \mathbb Z_{\frak p}$?

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  • $\begingroup$ For $x\in\mathcal O_K$, by definition, we have $x\mathcal O_K = \prod_{\mathfrak p} \mathfrak p^{\nu_{\mathfrak p}(x)}$, with $\nu_{\mathfrak p}(x) \geq 0$, that is, $|x|_{\mathfrak p} \leq 1$. $\endgroup$ – Thomas Poguntke Apr 25 '15 at 7:31
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Yes. As a DVR $\Bbb{Z}_{\mathfrak{p}}$ is integrally closed in its field of fractions $K$. It contains $\Bbb{Z}$ so it also contains the integral closure of $\Bbb{Z}$ in $K$ which is ...

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