I've been working on the problem of finding the maximal abelian extension of $\mathbb{Q}_5$ that is killed by $5$. In other words, find the abelian extension of $\mathbb{Q}_5$ with Galois group isomorphic to $\mathbb{Q}_5^\times/\mathbb{Q}_5^{\times 5}\simeq C_5\times C_5$. Now
$$(\mathbb{Q}_5^\times :\mathbb{Q}_5^{\times 5})=25$$
so essentially it's enough to find two disjoint $C_5$ extensions of $\mathbb{Q}_5$ and take their compositum. $C_5$ extensions don't seem too easy to construct, so by the local Kronecker-Weber theorem one could just look for cyclotomic extensions of degree divisible by $5$ and try to identify a subextension of degree $5$.
I've tried to find a good source about cyclotomic extensions of $p$-adics, but this seems to be a topic missing from all field theory books. Books on algebraic number theory seem to omit this topic too and our graduate algebra class didn't cover it either.
Now my intuition would guide me as follows. Working in $\mathbb{Q}$ one would immediately find $\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1})$ as a cyclic degree $5$ extension. We also have that $[\mathbb{Q}(\zeta_{25}):\mathbb{Q}]=20$, so that $[\mathbb{Q}(\zeta_{25}+\zeta_{25}^{-1}):\mathbb{Q}]=10$ and with $(\zeta_{25}+\zeta_{25}^{-1})^2=\zeta_{25}^2+2+\zeta_{25}^{-2}$ one could expect that setting $\alpha=\zeta_{25}^2+\zeta_{25}^{-2}$, we would have
$$[\mathbb{Q}(\zeta_{25}+\zeta_{25}^{-1}):\mathbb{Q}(\alpha)]=2\Rightarrow [\mathbb{Q}(\alpha):\mathbb{Q}]=5$$
(Note, I haven't actually checked if this first extension has degree $2$)
Finally, $\mathbb{Q}(\zeta_{25})\cap \mathbb{Q}(\zeta_{11})=\mathbb{Q}$, so we would have that
$$\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1},\alpha)/\mathbb{Q}$$
is a $C_5\times C_5$ extension of $\mathbb{Q}$. My question is then that does this work if we replace $\mathbb{Q}$ with $\mathbb{Q}_5$? More specifically:
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What would the Euler totient functions look like if defined as the order of the $n^\textrm{th}$ cyclotomic extension of $\mathbb{Q}_p$? If $p-1\mid n$, it would at least have to be different. Is $[\mathbb{Q}_5(\zeta_{25}):\mathbb{Q}_5]=[\mathbb{Q}(\zeta_{25}):\mathbb{Q}]$?
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Can we still expect to have e.g. $\mathbb{Q}_p(\zeta_n)\cap \mathbb{Q}_p(\zeta_m)=\mathbb{Q}_p(\zeta_{\gcd(n,m)})$? What about other local fields of characteristic $0$?
I haven't thought about these really at all as I would rather just find a good reference for this topic.
Best Answer
Books on algebraic number theory should have this.
For $n$ not divisible by $p$, the Galois group of $\mathbb{Q}_p(\zeta_n)$ over $\mathbb{Q}_p$ is the same as the Galois group of $\mathbb{F}_p(\zeta_n)$ over $\mathbb{F}_p$, basically because of Hensel's Lemma. But finite extensions of finite fields are cyclic, so it's easy to calculate the latter Galois group: It's cyclic of size equal to the multiplicative order of $p$ modulo $n$. The extension is also unramified.
For $n=p^k$, the Galois group of $\mathbb{Q}_p(\zeta_n)$ over $\mathbb{Q}_p$ is $(\mathbb{Z}/p^k \mathbb{Z})^{\times}$ , same as over $\mathbb{Q}$. For $k=1$, you can see directly that the cyclotomic polynomial $(x^p-1)/(x-1)$ is irreducible because it becomes an Eisenstein polynomial upon making the change of variable $x \mapsto x+1$. [Edit: This substitution doesn't always work for higher $k$; I'll have to think of another argument here.] The extension is totally ramified.
For general $n$, combine the previous 2 cases.