While reading my notes of a course in local class field theory, I arrived to a remark where it is said that given a complete discrete valuation field $K$, its maximal unramified extension
$$K^{ur}= \bigcup_{F / K \: fin. unr.} F $$
may not be complete. I was going to ask for a concrete example (that is, a Cauchy sequence in $K^{ur}$ that doesn't converge), but after some research in google I found one as an exercise in Local Fields and Their Extensions, by Ivan B. Fesenko, S. V. Vostokov:
Let $\pi \in K$ be a prime element, and let $k^{sep}$ be of infinite degree over $k$ (as in $K = \Bbb Q_p$, $k = \Bbb F_p$). Let $K_i$ be finite unramified extension of $K$, with $K_i$ strictly contained in $K_j$ for $i < j$. (We can do this in the above example because we have a 1.1 correspondence between finite unramified extensions of $\Bbb Q_p$ and finite extensions of $\Bbb F_p$.) Define
$$ \alpha_n := \sum_{i=1}^n \theta_i \pi^i $$
where $ \theta_i \in \mathcal{O}_{K_{i+1}} \setminus \mathcal{O}_{K_i}$. Show that $(\alpha_i)$ is a Cauchy sequence and that $\lim_n \alpha_n$ is not in $K^{ur}$.
Well, to show that it is a Cauchy sequence is trivial, and to see that the limit is not in $K^{ur}$ we argue like this: if it is in the union, it belongs to one of the $K_i$'s, but this contradicts the fact that $\alpha_j \notin K_j$ for $j > i$. Edit: The contradiction only appears once we fix representatives of $\mathcal{O}_{\widehat{K^{ur}}}$, see the nice counterexample by Torsten Schoeneberg below for details.
So here my question comes: how does the closure of $K^{ur}$ look like? Here an answer is given for $K = \Bbb Q_p$, but they just mention what it is and an explanation of this or an answer to my more general question will be welcomed.
Thank you!
Best Answer
This is a natural question, because it’s really easy to get overwhelmed by the situation. In the case of the completion of the maximal unramified of a local field $k$, here’s the way that I look at things: you have the maximal unramified extension, which I’ll call $K$, an infinite algebraic extension gotten by adjoining the $(p^n-1)$-th roots of unity for all $n$. Let’s call $\mathcal O$ the integers of $K$.
Now for the completion, $\overline K$: you can think of the elements of the integers there as series $\sum_ia_i\pi^i$, where each $a_i$ is in $\mathcal O$ and where $\pi$ is a chosen prime element of $k$. This representation isn’t unique. If you want a unique representation, restrict the $a_i$ all to be roots of unity of the type I mentioned above, or zero (these are the “Teichmüller representatives”).
If you start thinking about the completion of the algebraic closure of $k$, things get really confusing, partly because there’s no unique representation of an element there. But the first description in the paragraph above works in that case just as well.