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

Question

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!

Answer 1

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.

Answer 2

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).

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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 $\lim_{n \to \infty} \alpha_n = 17p \in \mathbb Z \subset K^{ur}$.

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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.

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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.)