# Why $\mathbb{Q}_p^{ur} \neq \widehat{\mathbb{Q}_p^{ur}}$?

Why is $\mathbb{Q}_p^{ur}$ not complete?

And is there a criterion to know when $K^{ur} = \widehat{K^{ur}}$ ? (where $K$ is a p-adic field, i.e. a field of characteristic 0 that is complete with respect to a discrete valuation that has a perfect residue field of characteristic $p > 0$. )

• Do you understand why $\overline{\mathbb{Q}_p}$ is not complete? Sep 11, 2014 at 16:46
• I know a proof of that where we look at $\sum p^{u_n}, u_n=3^{n^2}/2^{n^2}$
– gpst
Sep 11, 2014 at 17:01
• I find a proof of $\mathbb{Q}_p^{ur} \neq \widehat{\mathbb{Q}_p^{ur}}$ in Gouvea, p-adic numbers an introduction.
– gpst
Sep 11, 2014 at 17:16
• Then I don't think I understand your first question. Are you looking for help in understanding the proof? Or perhaps some conceptual argument as to why one would expect this? Sep 11, 2014 at 18:04
• I ask too quickly for the first question actually, but I don't find anything for the second part.
– gpst
Sep 12, 2014 at 9:27

Even though you say that you’ve found an explanation in Gouvêa’s book that $\mathbb Q_p^{ur}$ is not complete, let me give one here, for completeness.
The resideu field $\mathbb F_p$ of $\mathbb Z_p\subset\mathbb Q_p$ has an infinite tower of extensions, $\mathbb F_p=k_0\subset k_1\subset k_2\subset\cdots$, eacb=h inclusion being proper. There are many such towers, of course. EAch of the fields $k_n$ is a simple extension of $\mathbb F_p$, say generated by $\alpha_n$, a primitive $s_n$-th root of unity of some order. Each of these $\alpha_n$’s may be lifted (uniquely) to a primitive $s_n$-th root of unity $\zeta_n$, necessarily in the maximal unramified extension of $\mathbb Q_p$. Now take the Cauchy series $\sum_np^n\zeta_n$, certainly in the completion, and not in any finite extension of $\mathbb Q_p$, thus not in the maximal unramified (since that’s an algebraic extension of $\mathbb Q_p$). [I’m confident that this argument is not as clear or as economical as Gouvêa’s.]
Now: take a complete valued field $K$ of characteristic zero whose residue field $\kappa$ has characteristic $p>0$. If $\kappa$ is algebraically closed, then $K$ is already its own maximal unramified. If $\kappa$ is not algebraically closed, however, then it has an infinite ascending chain of algebraic extensions just as in the paragraph above, and the rest of the proof there goes through for this situation, so that the maximal unramified of $K$ is not complete.
In other words, the only way for $K^{ur}$ to be complete is for $K$ to be equal to $K^{ur}$.
• First, in the second last paragraph, seemingly you wanted to say that if $\kappa$ is separably closed. Next, I don't understand the existence of an infinite ascending chain of finite separable extensions. Note that $[\mathbb C:\mathbb R]=2$, there should be some characteristic-dependent argument involved. Apr 29, 2021 at 17:29
• Undoubtedly, @Yai0Phah , I had in mind the (special?) case that $\kappa$ was algebraic over the prime field, so your first point is well taken. The case that an algebraically closed field may be finite over a subfield is, of course, a characteristic-zero phenomenon. But it’s quite possible that my claim that there would be a proof “just as in the paragraph above” would fail if the residue field was not of the type mentioned at the beginning of this comment. Thanks for looking closely! Apr 29, 2021 at 17:56
• By the way, there is another interesting fact (which I don't see in the classical literature): if we take the topology on an algebraic extension $E$ of $\mathbb Q_p$ as the colimit topology of all finite intermediary extensions with canonical topologies (which is, in my eyes, more natural than the norm topology), then it is Cauchy-complete. A reference: Bhatt-Scholze, the pro-étale topology for schemes, Example 7.1.7. Apr 30, 2021 at 18:30
• I'd just like to point out that the fact that the limit of the Cauchy series $\sum p^n \zeta_n$ does not lie in any finite extension of $\mathbb Q_p$ has a short and wrong proof, which even made it into an otherwise reputable source cited in the near-duplicate math.stackexchange.com/q/1176495/96384 of this question. It of course also has a little more intricate, correct proof which you allude to in your answers here and there, but over there I felt the need to spell it out, together with a refutation of the short wrong proof, in a separate answer. Oct 12, 2021 at 4:57