0
$\begingroup$

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 $Q_p^{nr}$.

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

$\endgroup$
  • $\begingroup$ $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 '20 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 '20 at 21:55
  • 1
    $\begingroup$ 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 '20 at 21:59
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '20 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 '20 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.