Maximal tamely ramified abelian extension of $\mathbb{Q}_p$ is finite over the maximal unramified extension $\mathbb{Q}_p^{nr}$?

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 $$Q_p^{nr}$$.

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

• $Gal(Q^{nr}_p(p^{1/(p^\infty-1)}/Q^{nr}_p)$ is abelian not $Gal(Q^{nr}_p(p^{1/(p^\infty-1)}/Q_p)$ Jun 19, 2020 at 21:47
• Isn't $Gal(Q_p^{nr}/Q_p)$ abelian? and a tower of abelian extensions is abelian? Jun 19, 2020 at 21:55
• 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 $x\to ax+b,Z/(4)\to Z/(4)$) Jun 19, 2020 at 21:59

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

• 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. Jun 19, 2020 at 18:16
• $Q_3(i,3^{1/4})/Q_3$ is Galois but not abelian. And $Q_3(3^{1/4})/Q_3$ is not Galois. Jun 19, 2020 at 19:18

[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 $$K$$.