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It is well-known what the $p^n$-cyclotomic extensions (i.e., adjoining $p^n$-th roots of unity) of $\mathbb{Q}_p$ are (see Serre, Local fields for instance).

However, assume now that $K/\mathbb{Q}_p$ is an arbitrary finite extension. What can now be said about the $p^n$-cyclotomic extensions of $K$?

It is clear that this is a harder problem and I haven't been able to find any literature on this but I'm sure that it's out there. At least some cases (for instance, the case $K/\mathbb{Q}_p$ unramified should be similar to the "classical case" I think).

Edit: I should have specified what I mean by "What can be said...?". What I'm interested in in particular is the ramification groups and the jumps in the ramification filtration.

/Daniel

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    $\begingroup$ What do you mean "what can be said"? If you add the $p^n$-th roots of $1$ for all $n$, (and not just the $p$-th roots as in your question) you get an extension whose Galois group is an open subgroup of $Z_p^\times$, and whose residue field is a finite extension of $F_p$. These extensions are used a lot in $p$-adic Hodge theory, so I suggest you look at papers in that area, eg Fontaine's "Arithmétique des représentations galoisiennes $p$-adiques". There's a lot of info about the ramification of that extension, for example. $\endgroup$
    – Laurent Berger
    Commented Jul 8, 2013 at 7:55
  • $\begingroup$ @Laurent: I have edited the question to specify what I mean. I knew that there was a lot of information i $p$-adic Hodge theory concerning this but but my impression is that this is mainly concerning the infinite cyclotomic tower, not its finite constituent. I haven't seen that paper by Fontaine, but I'll have a look. $\endgroup$
    – Daniel Larsson
    Commented Jul 8, 2013 at 10:00
  • $\begingroup$ Typical that that Fontaine's paper is in Asterisque which I don't have access to. Do you have any other reference? $\endgroup$
    – Daniel Larsson
    Commented Jul 8, 2013 at 10:04
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    $\begingroup$ @Larsson: "Typical"? :-( See math.u-psud.fr/~biblio/pub/2000/abs/ppo2000_24.html for the preprint version of Fontaine's paper, and see mathoverflow.net/questions/96257/… concerning the availability of Asterisque. $\endgroup$
    – Laurent Berger
    Commented Jul 9, 2013 at 7:08
  • $\begingroup$ @Laurent: By "typical" I meant "Just my luck that I don't have access to..." For some idiotic reason it didn't occur to me that there was a preprint version. Thank you! $\endgroup$
    – Daniel Larsson
    Commented Jul 9, 2013 at 8:59

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Here is an easy example of a $K$ such that $K(\zeta_{p^2})$ is the unramified extension of $K$ of degree $p$.

Start with $F=\mathbf{Q}_p(\zeta_p)$, and consider the $\mathbf{F}_p$-space $F^\times/F^{\times p}$ lines in which correspond to degree-$p$ cyclic extensions of $F$. There are two special lines, the one (call it $C$) generated by the image of $\zeta_p$, and the one (call it $U$) such that $F(\root p\of U)$ is the unramified degree-$p$ extension of $F$. [One can say precisely which line $U$ is, but never mind.]

These two lines are distinct because $F(\root p\of C)=\mathbf{Q}_p(\zeta_{p^2})$ is totally ramified over $\mathbf{Q}_p$ (as you know), whereas $F(\root p\of U)$ is not, by the definition of $U$.

So the plane $CU$ contains at least one more line $D$ (distinct from $C$ and $U$), and the extension $K=F(\root p\of D)$ is a ramified degree-$p$ extension of $F$.

I claim that $K(\zeta_{p^2})$ is the unramified extension of $K$ of degree $p$, as you can easily verify by computing its ramification index and residual degree over $F$.

The special case $p=2$ gives the classic example : $K(\sqrt{-1})$ is the unramified quadratic extension of $K=\mathbf{Q}_2(\sqrt3)$.

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  • $\begingroup$ Hmm, so what you're saying is that not all $p^n$-extensions of $K$ are ramified? Maybe I should have known this. $\endgroup$
    – Daniel Larsson
    Commented Jul 9, 2013 at 9:02
  • $\begingroup$ @Daniel: Well, I do not know what you mean by $p^n$-extension, but if mean "extension of degree $p^n$" you might know that for every $p$-adic field, hence also for $\mathbb{Q}_p$, there exists a unique extension of degree $d$ which is unramified, and for all $d\geq 1$ (it is cyclic, moreover). This is in Serre's book. $\endgroup$
    – Filippo Alberto Edoardo
    Commented Aug 7, 2013 at 16:46
  • $\begingroup$ @FilippoAlbertoEdoardo: Well, I actually meant $p^n$-cyclotomic extensions. Yup, I knew about the result about unramifiedness (?), but I'm actually interested in wild ramification. $\endgroup$
    – Daniel Larsson
    Commented Aug 7, 2013 at 18:15
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    $\begingroup$ @Daniel: Well, in that case if $K/\mathbb{Q}_p$ is finite then $K(\zeta_{p^n})/K$ is wildly ramified for $n\gg 0$ but as Dalawat wrote it might very well be possible that for small $n$ the extension is unramified. $\endgroup$
    – Filippo Alberto Edoardo
    Commented Aug 8, 2013 at 1:46
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    $\begingroup$ @YijunYuan Consider the ramification index $e=e(K(\zeta_{p^n})/\mathbb{Q}_p)$: it can be written as a product in two ways: $e=e(K(\zeta_{p^n})/K)\cdot e(K/\mathbb{Q}_p)$ and $e=e(K(\zeta_{p^n})/\mathbb{Q}_p(\zeta_{p^n})\cdot e(\mathbb{Q}_p(\zeta_{p^n})/\mathbb{Q}_p)=e(K(\zeta_{p^n})/\mathbb{Q}_p(\zeta_{p^n})\cdot (p-1)p^{n-1}$. Since the power of $p$ that divides $e(K/\mathbb{Q}_p)$ is bounded independently of $n$ this shows that for $n\gg 0$, $p$ must divide $e(K(\zeta_{p^n})/K)$. $\endgroup$
    – Filippo Alberto Edoardo
    Commented May 10 at 18:08

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