Cyclotomic extension of p-adic fields

Question

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

Answer 1

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 $K=\mathbf{Q}_2(\sqrt3)$.