Let $p$ be a prime number and $K$ a finite extension of $\mathbb{Q}_p$. Put $K_\infty = K(\mu_{p^\infty})$, the field extension obtained by adjoining all $p$-power roots of unity to $K$.

I want to prove that: if $u \in {\mathcal{O}_K^\times}$ is not a root of unity, then the field extension $K_u$ of $K_\infty$ generated by the $p$-power roots of $u$ is non-abelian over $K$.

Here is a vague idea that I believe will allow us to prove the statement:

First, note that $\mathcal{O}_K^\times = F \times (1 + m_K)$, where $F$ is a finite group and $1 + m_K$ is the group of principal units. We may then reduce the problem to the case when $u$ is a principal unit. Kummer theory provides us with an isomorphism $$ K^\times / (K^\times)^{p^n} \cong H^1 (G_K, \mu_{p^n}), $$ for all $n \in \mathbb{Z}_{\geq 1}$. Taking the projective limits for $n$, we get $$\widehat{K^\times} \cong H^1 (G_K, \mathbb{Z}_p(1)).$$ We consider $u$ now as an element of $\widehat{K^\times}$. I saw in a recent paper by Khare and Wintenberger entitled, Ramification in Iwasawa theory and Splitting Conjectures, that

  • $K_u$ is the extension of $K_\infty$ corresponding to the fixed field of the kernel of the homomorphism arising from the image of $u$ under the map $$ f: \widehat{K^\times} \cong H^1 (G_K, \mathbb{Z}_p(1)) \rightarrow \text{Hom} (G_{K_\infty}, \mathbb{Z}_p)(1)^{\text{Gal}(K_\infty/K)}.$$

Let $\rho = f(u)$. Then, to prove that $K_u$ is non-abelian over $K$, it is enough to verify that the image of $\rho$ is non-abelian.

Now, with this line of argument,

(1) I do not know how the assumption that $u$ is a principal unit and is not a root of unity comes to use;

(2) I am having difficulty in understanding the statement in bullet, so I would be glad if someone can explain to me how $f$ was constructed and how come $K_u$ can be interpreted in the manner described above;

(3) I wonder if there is some effect to the proof if $K$ is assumed to contain some $p$-power roots of unity.

Of course, I also welcome other ideas that can lead to the proof of the above statement. Thanks in advance.

  • $\begingroup$ Is $K$ local here, or at least a non-Archimedean valued field? If not, what is $\mathscr{O}_K$? $\endgroup$ Jul 17, 2013 at 12:44
  • $\begingroup$ Apologies for the confusion. Although I am more interested in a more general case, I will now consider $K$ to be a finite extension of $\mathbb{Q}_p$. This will also enable me to understand what happens in a basic case. $\endgroup$
    – Octobris
    Jul 17, 2013 at 12:51
  • 2
    $\begingroup$ I think $Hom(G_{K_\infty},\mathbb{Z}_p)(1)^{Gal(K_\infty/K)}$ should actually be $Hom(G_{K_\infty}, \mathbb{Z}_p(1))^{Gal(K_\infty/K)}$. The map comes from inflation-restriction. Any nonzero element of the image gives a non-abelian extension as long as $K\neq K_\infty$. I will try to write a more complete answer a bit later if no one else does. $\endgroup$ Jul 17, 2013 at 20:55

1 Answer 1


Well, let me try in an elementary way. Pick any field $K$ of characteristic prime to $p$ and suppose it does not contain any $p^n$-th root of unity. Let $a\in K^\times\setminus (K^\times)^p$. Then, the extension $K(\sqrt[p^n]{a},\mu_{p^n})/K$ is normal, not abelian and its Galois group is isomorphic to $\Delta_n\ltimes\mathbb{Z}/p^n$ where $\Delta_n=\mathrm{Gal}(K_n(\zeta_{p^n})/K)$. To see this, start by observing that it is clearly Galois (it contains all roots $\xi\;\sqrt[p^n]{a}$ of $X^{p^n}-a$, for $\xi$ running in $\mu_{p^n}$), so the point is to study its Galois group. Set $F_n=K(\zeta_{p^n})$ and let $H=\mathrm{Gal}\big(F_n(\sqrt[p^n]{a})/F_n)\big)$ and $G=\mathrm{Gal}\big(F_n(\sqrt[p^n]{a})/K(\sqrt[p^n]{a})\big)$. I claim that

(*) the order of $a$ in the quotient $F_n^\times/(F_n^\times)^{p^n}$ is exactly $p^n$ (at least if $K/\mathbb{Q}_p$ is finite)

Admitting the claim, $H$ is a cyclic group of order $p^n$ (let $\eta$ be a generator) while $G$ is a cyclic group of order $d>1$ for some $d\mid p^{n-1}(p-1)$ (this is classic, see for instance Lemma 1 in Birch's paper in Algebraic Number Theory by Cassels-Frölich). Define a $\mathbb{Z}_p^\times$-valued character $\omega:G\to\mathbb{Z}_p^\times$ by $g(\zeta)=\xi^{\omega(g)}$ for all $\xi\in \mu_{p^n}$ and $g\in G$: this implies $g(\xi\;\sqrt[p^n]{a})=\xi^{\omega(g)}\;\sqrt[p^n]{a}$. Finally, there is a $j$ such that $\eta(\sqrt[p^n]{a})=\zeta_{p^n}^j\;\sqrt[p^n]{a}$ where $\zeta_{p^n}$ is a fixed generator of $\mu_{p^n}$. You can readily compute that for each $g\in G$, $$ g\eta(\zeta_{p^n}\sqrt[p^n]{a})=\zeta_{p^n}^{(1+j)\omega(g)}\;\sqrt[p^n]{a} $$ while $$ \eta g(\zeta_{p^n}\sqrt[p^n]{a})=\zeta_{p^n}^{\omega(g)+j}\;\sqrt[p^n]{a} $$ which shows at once that $g$ and $\eta$ do not commute unless $g=1$ and that $H\cap G=\{1\}$, namely the extension $F_n(\sqrt[p^n]{a})/K$ is not abelian with Galois group $G\ltimes H$, isomorphic to $\Delta_n\ltimes \mathbb{Z}/p^n$.

We are left with my claim (*): this is where one needs some Kummer theory. Indeed, consider the inflation-restriction sequence $$ 1\to H^1(\Delta_n,\mu_p)\to H^1(K,\mu_p)\to H^0(\Delta_n,H^1(F_n,\mu_p))\to H^2(\Delta_n,\mu_p) $$ and observe that $\Delta_n$-cohomology of $\mu_p$ vanishes: this can be seen by computing $H^2=\hat{H}^0$ which is trivial because $\mu_p(K)=\{1\}$, and then using that the Herbrand quotient of a finite module is $1$: finally, identify the $H^1$'s with the quotients by $p$-th powers by Kummer theory to find $$ H^0(\Delta_n,F_n^\times/(F_n/^\times)^p)\cong K^\times/(K^\times)^p. $$ In particular, we see that $a$ does not become a $p$-th power in $F_n^\times$ and since $F_n/\mathbb{Q}_p$ is finite we know $F_n^\times/(F_n^\times)^{p^n}$ is isomorphic to finitely many copies of $\mathbb{Z}/p^n$, so not being a $p$-th power coincides with having order $p^n$.

Now, back to your situation, you simply observe that the extension you call $K_u$ is the direct limit of extensions $F_n(\sqrt[p^n]{u})$ so the Galois group $K_u/K$ is the inverse limit of $\Delta\ltimes(\mathbb{Z}/p^n)$ none of which is abelian, so it is a non-trivial semi-direct product $$ \Delta\ltimes\mathbb{Z}_p $$ for some finite-index subgroup $\Delta\subseteq\mathbb{Z}_p^\times$, all this provided that $u\notin (K^\times)^p$. This is certainly true if $u$ is a generator of principal units and certainly false if it is a root of unity of order prime to $p$. Of course, if $u$ is a non-trivial principal unit in $(K^\times)^{p^t}\setminus (K^\times)^{p^{t+1}}$, say $u=v^t$ you repeat the above argument with $v$ instead of $u$ (may be, getting some headache due to index-shifting), and similarly if $K$ contains some $p^k$-th root of unity.

As for Khare-Wintenberger's argument, as Kevin observed, they are simply restating the above elementary computation expressing it in terms of cohomology classes but I guess nothing new appears (observe we used Kummer isomorphism in proving (*) and that, is all is needed).

  • $\begingroup$ Ooops. Sorry about the first line. What I meant was that $K$ is a field of characteristic not equal to $p$. Thanks for your answer. So, upon reading your answer, it seems to me that it doesn't matter weather $u$ is a unit that is not a root of unity or not. The extension will be non-abelian as long as it is not a $p$th power. Moreover, what about when $K$ contains some $p^n$th root of unity? $\endgroup$
    – Octobris
    Jul 17, 2013 at 11:54

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