The maximal unramified extension of a local field may not be complete

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!

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.

TL, DR: One claim cited in the question is actually wrong in the generality it is stated there, and some short arguments given for it are flawed; however, the issue can be fixed in sufficient special cases if one invokes more structure theory (as alluded to in Lubin's answer, which is, of course, entirely correct).

1. The exercise quoted in the OP is exercise 1a to chapter II, section 3 in Fesenko/Vostokov, page 55 here (they use $F$ where we use $K$). But part of what is claimed there, namely, that

for $ \alpha_n := \sum_{i=1}^n \theta_i \pi^i$ (where $ \theta_i \in \mathcal{O}_{K_{i+1}} \setminus \mathcal{O}_{K_i}$), we have $\lim_{n \to \infty} \alpha_n \notin K^{ur}$

is actually wrong unless one is more restrictive about what $\theta_i$ are allowed.

Example. Since $17$ is my favourite number, I will show an example where $\lim_{n \to \infty} \alpha_n = 17 p$ which quite obviously is $\in K^{ur}$.

Indeed, let $K = \mathbb Q_p, \pi =p$, $K_i := \mathbb Q_p(\zeta_{p^i-1})$, and let

$$\theta_1 := 17 - p \zeta_{p^2-1}$$

and for $i \ge 2$,

$$\theta_i := \zeta_{p^{i}-1} - p \zeta_{p^{i+1}-1}.$$

Then $\theta_i \in \mathcal{O}_{K_{i+1}} \setminus \mathcal{O}_{K_i}$ but we also have $$\alpha_n = 17p - p^{n+1} \zeta_{p^{n+1}-1}$$ for all $n$, so as claimed $\lim_{n \to \infty} \alpha_n = 17p \in \mathbb Z \subset K^{ur}$.

2. The issue with supposed arguments showing that the limit is $\notin K^{ur}$, like in the OP

if it is in the union, it belongs to one of the $K_i$'s, but this contradicts the fact that $\alpha_j \notin K_j$ [maybe $K_i$ is meant here, but it does not matter] for $j > i$.

is that while the first implication (that it belongs to some $K_i$) is true, and while it is also true that $\alpha_j \notin K_i$ for any $j \ge i$ (this holds true in my example in 1 too), these things just do not contradict each other. In a comment it is written

each $\alpha_i \in K_i \setminus K_{i-1}$, so when you make $i$ go to infinity, it will not be in any $K_i$. The limit exists, but it is not in any finite extension.

but these vague words contain an invalid conclusion from the terms of a sequence to its limit; as the example shows, we can very well have a sequence whose individual terms leave any finite extension, but that tells us more or less nothing about its limit.

I admit at first those claims sounded plausible to me too, but I also assume once one sees what the issue is, one sees how I made up my example, and how one can easily make many more.

3. How we can save what we want:

Of course e.g. Lubin is well aware of these things, but politely hides them a little in his answer, in the short remarks about uniqueness of certain representations. This idea does provide us with examples where the argument goes through. One just has to put a little more work into this, as in

Example: If in the situation described in 1 we instead let $$\theta_i := \zeta_{p^{i+1}-1},$$ then indeed $\lim \alpha_n \notin K^{ur}$.

Namely, the completion $\widehat{K^{ur}}$ is a field with a discrete valuation, uniformizer $p$, residue field $\mathbb F_p^{alg}$, and (by definition) complete. Hence, by general structure theory for such fields, once we fix a set of representatives $R \subset \mathcal{O}_{\widehat{K^{ur}}}$ of the residue field, then the standard series representation with such representatives is unique, i.e. if $\sum \beta_n p^n = \sum \gamma_n p^n$ and all $\beta_i, \gamma_i \in R$, then for all $n$ we have $\beta_n =\gamma_n$. But the joke is that we can choose a set of such representatives which is compatible with our tower of fields $K_i$; for example, letting $R_i := \mu_{p^i-1}$ (all roots of unity of order $p^i-1$) and $R:= \bigcup R_i$, now we can put the supposed arguments in 2 on solid ground: Namely, if $\alpha:=\lim \alpha_n \in K^{ur} = \bigcup K_i$, then it is in some $K_i$; but because $R_{i}$ is also a set of representatives of the residue field of the complete field $K_i$, this means there is also a representation

$$\alpha = \sum \gamma_n \cdot p^n \qquad \qquad (I)$$

with all $\gamma_n \in R_i$. But in $\widehat K^{ur}$, of course we still have

$$\alpha = \sum \zeta_{p^{n+1}-1} \cdot p^n \qquad \qquad (II).$$

Now because $\zeta_{p^{j+1}-1} \notin R_i$ for all $j \ge i$, but $R_i \subset R$, the expressions $I$ and $II$ contradict the uniqueness of series representations with coefficients in $R$ in the field $\widehat{K^{ur}}$.

(It's also fun to see where exactly this kind of proof breaks down in my example in 1. It happens not where one might first think: The series representations $\alpha = 17 p$ versus $\alpha = \sum \theta_i p^i$ do not contradict each other, because they do not share the same set of representatives as coefficients.)