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I am actually quite confused. I have done an exercise that $\mathbb{Z}_p$ is a completion of $\mathbb{Z}$ w.r.t. the $p$-adic norm. Then again I got to know after reading somewhere that $\mathbb{Z}_p$ is also a completion of $\mathbb{Z}_{(p)}$. So is the completion of a valuation ring of field $K$ related to the valuation ring of the completion of $K$?

Another question I wanted to ask. Whether any complete field (where Cauchy sequences converge) is always a completion over another non complete field?

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  • $\begingroup$ The completion $\widehat{R}$ of a valuation ring $R$ at its valuation is a valuation ring. The valuation extends to the fraction field $Frac(R)$ and the valuation ring of $\widehat{Frac(R)}$ is the same as $\widehat{R}$. Note that it is not the same as the $\mathfrak{m}$-adic completion $\varprojlim R/\mathfrak{m}^n$. $\endgroup$
    – reuns
    May 20, 2022 at 19:34
  • $\begingroup$ @reuns So they are same.Now if I define the valuation ring of $(K,|\quad|)$ as the set $R=\{x\in K| |x|\leq1\}$ and that of completion $(\widehat{K},||\quad||)$ as the set $R^{\prime}=\{x\in K| ||x||\leq1\} $Then how can I show that $R^{\prime} = \widehat{R}$? Any hints or anything.I dont know actually what is m-adic completion. $\endgroup$
    – No pie
    May 21, 2022 at 2:22

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I think the $K$ in your definition of $R'$ should be $\hat{K}$. Clearly, $\hat{R} \subset R'$. Now take an element $a \in R'$, say, $a$ is represented by a Cauchy Sequence $(a_n)$ such that $|a_n|$ converge to $1$. If $|a| < 1$, then there exists an integer $N$ such that for $n \geq N$, $|a_n|$ is less than $1$, then $a$ is represented by $(a_{n+N}) \in \hat{R}$ and we win. Otherwise, $|a|=1$, then $a$ is represented by $(a_n/|a_n|) \in \hat{R}$ and we win!

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