Ring of integers in a field of fractions 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 ? 
 A: 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 $Q$. 
A: Here is a counterexample: Let $k$ be a field and let $R=k[[X,Y]]$ be the ring of formal power series with coefficients in $k$. Let $Q=k((X,Y))$ be its field of fractions. For $F\in k[[X,Y]]$, $F\neq 0$ define $v(F)$ as the least degree of a monomial appearing in $F$, and extend $v$ to $k((X,Y))$ by $v(F/G)=v(F)-v(G)$. Then $v$ is a valuation in $k((X,Y))$ that induces a non-archimedian absolute value in $k((X, Y))$, namely $|F|=2^{-v(V)}$. It is well-known that $R$ is complete, but I will show that $Q$ is not. This clearly implies that $O_Q$ is neither complete.
A specific instance of Cauchy non-convergent sequence is the series $\sum\limits_{n=0}^\infty \frac{X^{n^2}}{Y^{n^2}}Y^n$.
Call $T=X/Y$ and consider the map $\phi:k[[X,Y]]\longrightarrow k(T)[[Y]]$ given by
$$
\phi(\sum_{n=0}^\infty F_n)=\sum_{n=0}^\infty F_n(T,1)Y^n,
$$
where $F_n$ is the sum of all monomials of $F$ of degree $n$.
For $F\in k(T)[[Y]]$ let $v(F)$ be the least degree of a monomial appearing in $F$. This induces a valuation on the quotient field $k(T)((Y))$, which induces a complete non-archimedian absolute value (it is complete because now we have just one indeterminate).
Considering this absolute value, $\phi$ is clearly an isometry (not necessarily onto), and it extends to an isometry $\phi: k((X,Y))\longrightarrow k(T)((Y))$.
If we identify $R$ with its image, it consists of all power series $\sum p_n(T)Y^n$ such that $\deg p_n(T)\leq n$. Let us identify also $K$ with its image in $k(T)((Y))$. Notice that $k(T)\subset K$, since, if $p(t)/q(T)\in k(T)$ and $n$ is greater than the degree of both $p$ and $q$, we have
$$
\frac{p(T)}{q(T)}=\frac{p(T)Y^n}{q(T)Y^n}\in K.
$$
Since also $Y\in K$, we have in fact $k(T)(Y)\subset K\subset k(T)((Y))$. But $k(T)(Y)$ is dense in $k(T)((Y))$. So, if $K$ is complete, and hence closed, it must be $K=k(T)((Y))$. Let us show that this is not the case.
Consider $w:k[T]\longrightarrow \mathbb Z$ the valuation given by $w(p)=-\deg p$, which can be extended to $w:k(T)\longrightarrow \mathbb Z$.
Every $F\in R$ can be expressed as $F=Y^N\sum_{n=0}^\infty p_n(T)Y^n$, with $p_0\neq 0$ and $\deg p_n\leq n+N$. Put $\deg p_0=M\leq N$.
It is well-known that $\sum_{n=0}^\infty p_n(T)Y^n$ is a unit in $k(T)[[Y]]$. Consider
$$
(\sum_{n=0}^\infty p_n(T)Y^n)^{-1}=\sum_{n=0}^\infty q_n(T)Y^n,
$$
with $q_n(T)\in k(T)$. Computing inductively each $q_n$ it is easily seen that
$$
w(q_n)\geq M-n(N-M+1).
$$
Now, let $G=\sum_{n=0}^\infty r_nY^n\in R$ be an arbitrary element. We know that $w(r_n)\geq -n$. Then
$$
F^{-1}G=Y^{-N}\sum_{m=0}^\infty\sum_{n=0}^m q_nr_{m-n}Y^m=Y^{-N}\sum_{m=0}^\infty p_mY^m,
$$
and again it is easily seen that
$$
w(p_m)\geq M-m(N-M+1).
$$
Hence, for each $S\in K$, there exist $M$, $N\in \mathbb N$, with $M\leq N$ such that
$$
S=F^{-1}G=Y^{-N}\sum_{m=0}^\infty p_mY^m,\qquad w(p_m)\geq M-m(N-M+1).
$$
Hence, it is clear that
$$
S=\sum_{m=0}^\infty T^{m^2}Y^m\notin K.
$$
On the contrary,
$$
m^2\leq -M+m(N-M+1),
$$
for all $m$, but this is impossible.
