Here all the extensions are finite.

In local class field theory, one learns that the abelian extensions $\mathbb{Q}_{p}(\zeta)/\mathbb{Q}_{p}$, where $\zeta$ is a $p^{n}$-th root of unity, are in bijective correspondence with the subgroups $(p)\times (1+p^{n}\mathbb{Z}_{p})$ of the multiplicative group $\mathbb{Q}_{p}^{\times}$.

Do we know similar descriptions for other 'types' of extensions of $\mathbb{Q}_{p}$?

For example, we can play around and wonder what kind of abelian extensions $L_{n}/\mathbb{Q}_{p}$ correspond to the subgroups \begin{equation} (p^{2})\times (1+p^{n}\mathbb{Z}_{p}) \end{equation} and we can ask this for many other subgroups. Also, I wonder if we can detect ramification looking at the these subgroups.

For example, it is known that the $p^{n}$-th cyclotomic extensions I mentioned above are totally ramified, and just so happens that the "$(p)$"-part of the corresponding subgroup has exponent $1$.

I did this for a couple of examples, and it looks like an extension corresponding to some subgroup of the form \begin{equation} (p^{f})\times H, \end{equation} for some subgroup $H$ of $\mathbb{Z}_{p}^{\times}$, has inertia degree $f$. Is this true in general? Thanks a lot.

  • $\begingroup$ First of all--you should put FINITE abelian extension. Also, I have no idea how you got the original claim. Local class field theory says that the finite abelian extensions of $\mathbb{Q}_p(\zeta_{p^n})$ correspond to the finite index subgroups of $\mathbb{Q}_p(\zeta_{p^n})^\times$ which is $(1+\zeta_{p^n})^\mathbb{Z}\times \mathbb{Z}_p[\zeta_{p^n}]^\times$ and the latter is isomorphic to $(\Z/p^n\Z)\times (\Z/(p-1)\Z)\times \mathbb{Z}_p^{p^{n-1}(p-1)}$. Am I missing something? PS, your last question should be answered, I assume, by the fact that the upper numbering filtration $\endgroup$ – Alex Youcis Nov 5 '16 at 10:32
  • $\begingroup$ on inertia maps to the usual Lie filtration on the units of your valuation ring. $\endgroup$ – Alex Youcis Nov 5 '16 at 10:33
  • $\begingroup$ Hey, thanks for commenting. Don't know if I understand, though. I'm taking $\mathbb{Q}_{p}^{\times}$ as the base field and applying local CFT, so finite abelian extensions correspond to closed finite index subgroups of $\mathbb{Q}_{p}^{\times}$. In Neukirch's book ANT, there is a proposition (1.8; page 323) saying that these $p^{n}$-th cyclotomic extensions of $\mathbb{Q}_{p}$ correspond to the subgroups of the form $(p)\times U^{(n)}$. Also, I think your final sentence may be missing a word. The fact that what? $\endgroup$ – Shoutre Nov 5 '16 at 15:30
  • $\begingroup$ Did you see my second comment? But you're not alking about finite abelian ecxtensions of $\mathbb{Q}_p$, right? Finite abelian extensions of $K$ correspond to finite index subgroups of $K^\times$. Why then for $K=\mathbb{Q}_p(\zeta_{p^n})$ are you taking $\mathbb{Q}_p^\times$ and not $K^\times$? $\endgroup$ – Alex Youcis Nov 5 '16 at 21:52
  • $\begingroup$ OH, I misread a single word--I thought you said "abelian extensions OF $\mathbb{Q}_p(\zeta)/\mathbb{Q}_p$". OK, now I understand--sorry. Anyways, for any finite index subgroup of $\mathbb{Q}_p^\times$ you can always explicitly show it contains one of the form $(p)\times(1+p^n\mathbb{Z}_p)$. If you can then find that the index between your finite index subgroup and this group is, say, $G$ then your extension is just $\mathbb{Q}_p(\zeta_{p^n})^G$. Also, I still think my second comment about the inertia filtration answers your second question--does it make sense now? $\endgroup$ – Alex Youcis Nov 5 '16 at 21:54

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