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Since I just started to learn algebraic number theory and feel quite insecure about the subject, I apologize in advance for any wrong notations or conclusions. I tried to write down what I understood and wanted to verify whether the following is correct or not:

Let $K= \mathbb{Q}$ and let $p$ be a prime number. If I understood correctly, the field of $p$-adic numbers $\mathbb{Q}_p$ is the completion of $\mathbb{Q}$ wrt. the prime ideal $(p) \subseteq \mathcal{O}_K = \mathbb{Z}$.

I think what this means is that $\mathbb{Q}_p = K_{v_{(p)}}$ where $K_{v_{(p)}}$ is the completion of $K$ wrt. $v$ where $v_{(p)} : K \to \mathbb{Z} \cup \{ \infty \}$ is the valuation with $v_{(p)}(x)$ for $x \in K$ defined as the exponent of $(p)$ in the factorization of $(x)$ in prime ideals, and $v_{(p)}(0) = \infty$.

Could you please tell me if my line of thought is correct?

ANT Learner
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    Yes, you can express it in terms of valuations like that. But personally I think it's easiest to think about in terms of the $p$-adic metric: consider the rational numbers as a metric space with distance function $d(x, y) = p^{-v_p(x - y)}$, that is, $x$ and $y$ are closer the more times their difference is divisible by $p$. Then $\mathbb{Q}_p$ is the completion of this metric space, in the same way $\mathbb{R}$ is the completion of $\mathbb{Q}$ with the usual metric $d(x, y) = \lvert x - y \rvert$. – Daniel Hast Jan 14 '21 at 16:38
  • @DanielHast: Isn't the completion wrt. a valuation not basically the same as the completion wrt. the absolute value (or the corresponding metric space) defined by this valuation? – ANT Learner Jan 14 '21 at 16:51
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    @ANTLearner completion with respect to the valuation and with respect to the corresponding absolute value are the same thing – D_S Jan 14 '21 at 18:45
  • $\Bbb{Q}p=\Bbb{Z}_p[p^{-1}]$. There are two equivalent constructions of the $p$-adic integers: either as the completion of $\Bbb{Z}$ for the $p$-adic metric ($|a|_p=p^{-v_p(a)}$ so it is the same as completion for the $p$-adic valuation), or as the projective limit $\varprojlim{k\to \infty} \Bbb{Z/p^k Z}$. We can refer to both by saying 'the ring of limits of sequences of integers that converge $\bmod p^k$ for all $k$'. – reuns Jan 14 '21 at 21:01

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