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I came across an curious exercise in Neukirch's algebraic number theory book. Exercise 2 page 176 (Chapter II section 9) asks the following

Prove that the maximal tamely ramified abelian extension V of $\mathbb{Q}_p$ is finite over the maximal unramified extension $\mathbb{Q}_p^{nr}$ of $\mathbb{Q}_p$.

I suspect that this is wrong because one can always consider the extensions

$$ L_n = \mathbb{Q}_p^{nr} (p^{\frac{1}{p^n - 1}}) $$

$L_n$ is tamely ramified and has degree $p^n - 1$ since $X^{p^n - 1} - p$ is irreducible in $\mathbb{Q}_p$ (by Eisenstein criterion).

Am I missing something here?

Here is why I think $L_n$ is an abelian extension.

Call $K_n = Q_p^{nr}(p^{1/n})$ for $n$ prime to $p$. The extension $K_n/Q_p$ (I think) is abelian for all $n$ prime to $p$ and here is why:

It suffices to show that $K_n/Q_p^{nr}$ is abelian (since $Q_p^{nr}/Q_p$ is already abelian).

Any $Q_p^{nr}$-linear automorphism $\sigma$ of $K_n$ is determined by the image $\sigma(p^{1/n})$.

The Galois group elements of $Gal(K_n/Q_p^{nr})$ are then the elements $\sigma_i$ with

$$\sigma_i(p^{1/n}) = p^{1/n} \zeta_{n}^{i}$$

for $i = 0, 1, \dots, n-1$ and these commute since $\zeta_n$ ( a primitive $n$-th root of the unit) is in $Q_p^{nr}$.

$$ \sigma_i \sigma_j(p^{1/n}) = \zeta_{n}^{i} \zeta_{n}^{j} p^{1/n}$$

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  • $\begingroup$ $Gal(Q^{nr}_p(p^{1/(p^\infty-1)}/Q^{nr}_p) $ is abelian not $Gal(Q^{nr}_p(p^{1/(p^\infty-1)}/Q_p) $ $\endgroup$
    – reuns
    Jun 19, 2020 at 21:47
  • $\begingroup$ Isn't $Gal(Q_p^{nr}/Q_p)$ abelian? and a tower of abelian extensions is abelian? $\endgroup$
    – yassine
    Jun 19, 2020 at 21:55
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    $\begingroup$ Of course no towers of abelian extensions aren't $Q(2^{1/4})/Q(2^{1/2})/Q$. And $Gal(Q(2^{1/4},i)/Q)$ is the dihedral group $D_8$ (same as group of affine transformations $x\to ax+b,Z/(4)\to Z/(4)$) $\endgroup$
    – reuns
    Jun 19, 2020 at 21:59

2 Answers 2

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$V$ is a tamely ramified extension of $Q_p$ and it contains $Q_p^{nr}=\bigcup_{p\ \nmid\ m} Q_p(\zeta_m)$ so $$V=\bigcup_j Q_p^{nr}(p^{1/n_j})$$ for some $p\nmid n_j$ and $n_j | n_{j+1}$.

Since any subextension of an abelian extension is abelian

It suffices to find for which $n$ we have $Q_p^{nr}(p^{1/n})/Q_p$ is abelian and tame.

ie. when $Q_p(p^{1/n})/Q_p$ is abelian and tame.

This happens iff $n | p-1$.

Whence $V=Q_p^{nr}(p^{1/(p-1)})$

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  • $\begingroup$ I don't see how $Q_p^{nr}(p^{1/n})$ can fail to be abelian when $n$ is prime to $p$. I tried to explain my argument in the edited version of the question. I should have made that clearer, my bad. $\endgroup$
    – yassine
    Jun 19, 2020 at 18:16
  • $\begingroup$ $Q_3(i,3^{1/4})/Q_3$ is Galois but not abelian. And $Q_3(3^{1/4})/Q_3$ is not Galois. $\endgroup$
    – reuns
    Jun 19, 2020 at 19:18
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[If $K\subset L\subset M$ are Galois extensions and $M/L, L/K$ are abelian, $M/L$ need not be abelian. In fact, if one has an exact sequence of groups $$1\to H\to G\to K\to 1$$ where $H$ is normal in $G$ and $H, K$ abelian, $G$ can well be non-abelian. ($G = S_3$, for example)]

As for the problem, let $K$ be a field that is complete with respect to a non-trivial discrete valuation $v$ and suppose the residue field is $\mathbb{F}_q$. Let $x$ be a uniformizer in $K$, $d=q-1$, $\mathcal{O}$ the valuation ring in $K$, and $U, \mathfrak{m}$ the group of units and the maximal ideal in $\mathcal{O}$. Denote by $K_d$ the extension $K_{nr}(x^{\frac{1}{d}})$. Since $K$ contains all the $d$-th roots of unity (by Hensel's lemma), $K(x^{\frac{1}{d}})$ is Galois, and is therefore abelian. Thus $K_d$ is abelian.

Let $L/K$ be an abelian tamely ramified extension. The local reciprocity map $\omega: K^*\to \mathrm{Gal}(L/K)$ maps $U$ to the inertia subgroup $I$ and $1+\mathfrak{m}$ to $1$. Then, $\omega$ maps the residue field surjectively to $I$. So $I$ is finite of order dividing $d=q-1$. One now has $L\subset K_d$, and $K_d$ is the maximal abelian unramified extension of $K$.

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