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 $j > i$.

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!

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    $\begingroup$ Could you explain what you mean by 'look like'? That explanation works well for $\mathbb{Q}_{p^r}$ as well. What do you want in general? Also, it's worth mentioning, if $K$ is local, and $L/K$ an extension. Then, $L$ is complete if and only if $L$ is a finite extension. This follows pretty easily from the Baire category theorem. $\endgroup$ – Alex Youcis Mar 5 '15 at 12:25
  • $\begingroup$ I'm asking for the completion of $K^{ur}$ wrt the absolute value induced by the one in $K$. You can always take the completion, and since $K^{ur}$ is not complete, we will get something different. What? This is what I'm asking. For general I want to start with a different complete discrete valuation field (for example of positive characteristic). And I don't understand how can you use he Baire category theorem to conclude that, could you expand it to an answer? Thank you! $\endgroup$ – Pedro A. Castillejo Mar 5 '15 at 13:03
  • $\begingroup$ I can write up an answer, but it won't be more satisfactory than Lubin's answer below (which, as I commented, is the same as the answer in the post you linked). As for the Baire category theorem, proceed as follows. Let $L/K$ be infinite, and for each finite subextension $F$, let $U_F$ be $L-F$. Show that each $U_F$ is dense, and open, but $\displaystyle \bigcap U_F=\varnothing$. But, if $L$ were complete, it would be a Baire space, but the above shows this is not true. $\endgroup$ – Alex Youcis Mar 6 '15 at 6:29
  • $\begingroup$ @PedroA.Castillejo can you explain why does the limit doesn't exist in the argument you have given, i get the fact that $\alpha_j \not \in F_I$ for all $ j >i$, but why does that imply the limit is not in $\alpha_i$. Also in the above comment how we show that $F$ is closed in L? $\endgroup$ – Chirantan Chowdhury Jun 17 '17 at 17:05
  • $\begingroup$ @PedroA.Castillejo can you explain why if $\alpha_j \not \in F_i$ for all $ j >i$ implies that the limit is not in $F_i$? $\endgroup$ – Chirantan Chowdhury Jun 17 '17 at 17:32

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.

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    $\begingroup$ Dear Lubin, I believe you're rewriting what was linked in the original post. That if $k$ is the residue field of $K$, the $\widehat{K^\mathrm{u.r.}}$ is $W(\overline{k})$. I agree that this is a reasonable description, but perhaps this is not what the OP is looking for. But, perhaps I am confused. $\endgroup$ – Alex Youcis Mar 6 '15 at 1:03
  • $\begingroup$ @AlexYoucis, you may be right. But I can not imagine a more perspicuous description. $\endgroup$ – Lubin Mar 6 '15 at 13:40
  • $\begingroup$ And of course, @AlexYoucis, if the base was a ramified field, then the maximal unramified is no longer described by the Witt construction. $\endgroup$ – Lubin Mar 7 '15 at 22:31
  • $\begingroup$ That's a good point. But, it seems that in your description, you are exactly describing $W(\overline{\mathbb{F}_q})$. Am I missing something? Thanks! $\endgroup$ – Alex Youcis Mar 7 '15 at 22:38
  • $\begingroup$ Yes, @AlexYoucis, because I took powers of $\pi$, not powers of $p$. Naturally (because $k$ is finite over $\Bbb Q_p$), what I described is the tensor product of $k$ with (the fraction field of) the Witt vectors, but it certainly is not the Witt-construction tout simple. $\endgroup$ – Lubin Mar 8 '15 at 18:49

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