# About cyclotomic extensions of $p$-adic fields

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:

1. 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 $[\mathbb{Q}_5(\zeta_{25}):\mathbb{Q}_5]=[\mathbb{Q}(\zeta_{25}):\mathbb{Q}]$?

2. 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.

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## 2 Answers

It sounds as if you are a graduate student trying to learn some algebra number theory. (I'm sorry if I've guessed wrong on this point.) In this case I would encourage you to try to solve this question yourself, rather than look for a reference.

One hint is to think about ramification:

You can make one (and only one) $C_5$-extension that is totally unramified. Such extensions are always cyclotomic extensions. (They are given by extensions of the corresponding residue fields, which are for finite fields are always cyclotomic.)

You can also find a $C_5$-extension which is totally ramified. This can also be taken to be cyclotomic. Which cyclotomic extensions will be totally ramified at $5$?

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Thanks, you guessed right I'm a grad student. I did actually construct the unramified extension e.g. take a root of X^5-X+1 (I used the functorial correspondence with separable extensions of the residue field). I just hoped that cyclotomic fields would be easier to work with as that is the canonical thing to do over $\mathbb{Q}$. – Edvard F Sep 25 '11 at 14:21
@Edvard F: Dear Edvard, Using cyclotomic extensions is sensible.  As I wrote in my answer, unramified extensions are cyclotomic (e.g. the degree $5$ extension of $F_{\mathbb 5}$ is obtained by adjoining a $3124$th root of $1$).  In general, as Ted wrote in his answer, unramified extensions are obtained by adjoining $\zeta_n$, for some appropriate choice of $n$ prime to $p$.  To understand the extensions obtained by adjoining $\zeta_{p^k}$, you can use the Eisenstein arguments that Ted mentions, or ramification arguments.  The latter are more general, flexible, and powerful, so if you are ... – Matt E Sep 25 '11 at 17:45
... interested in learning algebraic number theory, I recommend that you learn them.  Regards, – Matt E Sep 25 '11 at 17:45

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 degree $n/\gcd(n,p-1)$. 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.

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Thanks. The $n=p^k$ case was what caused me a headache, so seems like I need to think about. I also didn't find explicit mentioning of cyclotomic extensions of $p$-adics in Lang's or Neukirch's Algebraic Number Theory. I might have to look at Serre's Local Fields. – Edvard F Sep 25 '11 at 15:27
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