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Let $R$ be ring with complete non archimedian absolute value.

Let $Q$ be the associated field of fractions with the extended absolute value.

Does the ring $O_Q = \{x\in Q | |x|\leq 1\}$ is complete ?

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Any closed subspace of a complete metric space is complete –  jspecter Jan 15 '13 at 7:36
    
@jspecter: so you say Q is complete ? –  colge Jan 15 '13 at 7:46

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up vote 1 down vote accepted

This is not a complete answer but I think it is worth to be written down.

It is clear that $O_Q$ is complete if and only if $Q$ is complete. But in general I am convinced that $Q$ is not complete.

Consider an example. Let $k$ be a complete field with respect to a non-archimedean absolute value. Let $R=k\langle T\rangle$ be the ring of power series $\sum_{n\ge 0} a_nT^n\in k[[T]]$ such that $(a_n)_n$ tends to zero. (It is a so called Tate algebra or affinoide algebra). Let $$ ||\sum_n a_nT^n||=\max_n \{ |a_n|\}.$$ This is clearly a non-archimedean norm on $R$, and it is easy to see that the norm is multiplicative. So $R$ is a domain (it is even a Dedekind domain) and the norm induces an absolute value on $Q$. It is not hard to show $R$ is complete.

Now what is $Q$ ? There is a Weierstrass preparation theorem (you have to look at some books on rigid analytic geometry as that of Bosch-Güntzer-Remmert or Fresnel-van der Put) which shows that any non-zero element of $R$ can be factorized into $P(T)u(T)$ with $P(T)\in k[T]$ and $u(T)$ invertible in $R$. So every element of $Q$ is a fraction $f(T)/P(T)$ with $f(T)\in R$ and $P(T)\in k[T]$. In other words $Q=R\otimes_{k[T]} k(T)$. It it too "algebraic" to be complete.

Take a sequence $(a_n)_n$ in $k$ going to zero and consider the sequence $f_n(T)=\sum_{0\le k\le n} a_kT^{-k}\in Q$. This is a Cauchy sequence (note that $|T|=||T||=1$). It converges to $\sum_{n\ge 0}a_nT^{-n}\in k[[1/T]]$. It can happen that this power series belongs to $Q$ (e.g. $a_n=\pi^n$ for a fix $\pi\in k$ with $|\pi|<1$). But if the $a_n$ are sufficiently random, I can't see how $f_n(T)$ could converge in $Q$.

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